If x, y, z are all positive one and add up to 1 they each have to be less than 1 but then their inverses are each bigger than 1 so 1/x + 1/y + 1/z > 1 + 1 + 1 > 3. That shouldn't take more than 20 seconds
Simply substitute the sum x+y+z for the 1 in the numerator. You get: x+y+z=1 1/x+1/y+1/z>3 (x+y+z)/x+(x+y+z)/y+(x+y+z)/z>3 Perform term-by-term division: 1+(y+z)/x+1+(x+z)/y+1+(x+y)/z>3 (y+z)/x+(x+z)/y+(x+y)/z>0 Because x, y, z > 0, the inequality is true.
Intuitively, the symmetric case is x=y=z=1/3, which would yield 9 in the second inequality which is larger than 3. Any other combination of x,y,z will require one of them to be less than 1/3 and another to be more than 1/3. The one that's less than 1/3 alone makes the sum of the reciprocals bigger than 3.
Oh cool, some friend showed me problems like that I didn't see them before but all these 3 methods are things I intuitively guessed in the past, good to know I wasn't completely wrong
Yeah, that's how I did it also after considering doing the full partial derivatives to find the extremum. Before picking a piece of paper, I realized that the order 3 symmetry in x, y, z force the extremum to be in x=y=z, then you just check if you got a minimum or maximum, which is also trivial. But I did not know the chain of means inequality, which I will remember because it's a nice formalisation of intuitions I have used time and time again and each time did a quick verification. This will makes things faster and easier in the future 👍
Thank you for another wonderful and blessed video. Substituting x + y + z for 1 in each term in 1/x + 1/y + 1/z is so cool and produces a great solution almost immediately. I was familiar with the GM < or = AM inequality but had not come across those involving HM and QM before, so thank you for helping me to never stop learning. Another proof using just the GM-AM inequality is as follows. Let L = 1/x + 1/y + 1/z Case 1: x=y=z As in your solution, this means x=y=z=1/3 so L=3+3+3=9. Hence L>3. Case 2: x y and z not all equal Adding its 3 fractions, L = S/xyz (1) where S = yz+zx+xy By the GM-AM inequality, S/3 > cube root (yz*zx*xy) The inequality is strict because it is equal only when yz, zx and xy are all equal, which is not so in this case (because yz=zx means x=y and zx=xy means x=z, contradiction) Hence S > 3 * cube root (x^2*y^2*z^2) Observe that x, y and z are all > 0 and < 1. Hence x^2 > x^3 and similarly for y and z Hence S > 3 * cube root (x^3*y^3*z^3) = 3xyz = 3S/L from (1) Hence S > 3S/L. And S>0. Hence L>3. As others have noted, just 0 < x, y and z < 1 is enough to give 1/x, 1/y and 1/Z all > 1 and hence L>3. So the above proof is overkill but still fun for me to play with. Again, thank you for your lovely videos and i look forward to seeing the next one. God bless you ❤
It is actually quite easy to show that the inequality >= 9 (stronger than >3)… Since x+y+z=1 it follows that (1/x)+(1/y)+(1/z) = (x+y+z) *((1/x)+(1/y)+(1/z)). Expanding this we get (1/x)+(1/y)+(1/z) = 3 + (x/y) + (y/x) + (x/z) + (z/x) + (y/z) + (z/y). Now for any a>0, it is easy to show that (1/a)+a >= 2 with equality iff a=1. Applying this to the above with a=x/y, x/z, y/z we get (1/x)+(1/y)+(1/z) >= 3 + 2 + 2 + 2 = 9, with equality iff all numbers are equal to 1/3 . Regardless I enjoyed the video since it gave a few different approaches - drives the point home that there isn’t always a unique approach 😊
Very nice, thank you. Equivalent to replacing 1 with x + y + z in each term for the first step but then working on the fractions to establish a lower bound on their sum greater than simply 0.
Cruder but simpler and sufficient for the problem set is to apply a + 1/a > or = 2 to x, y and z directly to give 1/x+1/y+1/z > or = 2+2+2-(x+y+z) = 5 > 3.
The simplest proof for me: Given that x + y + z = 1 and x, y, z > 0, we know that each of x, y, and z must be less than 1 because they sum up to one. Therefore, 1/x, 1/y, and 1/z are each greater than 1 because x, y, and z are less than one. Consequently, their sum is strictly greater than 3 because each term is more than one and there are three terms.
How complicated can he make this? Each reciprocal is greater than 1.......QED. He clearly shows that he likes algebra, in most cases it is simply not needed.
@@normtyneships194 The algebra in the first step is pretty useful for someone who might not be strong in logic, showing that the sum is 3 + [a bunch of + terms] puts it in very bluntly that it's greater than 3
All three x, y, z are less than 1, therefore all three 1/x, 1/y, 1/z are greater than 1. So, the addition of those reciprocals will always be greater than 3. That’s it.
Another approach: It is clear that x < 1, y < 1, and z < 1. x < 1 implies that yz*x < yz (1) y < 1 implies that xz*y < xz (2) z < 1 implies that xy*z < xy (3) Adding (1), (2), and (3) side by side yields: 3xyz < yz+xz+xy Dividing both sides by xyz (which is positive) yields: 3 < (yz+xz+xy)/xyz = 1/x +1/y + 1/z Q.E.D.
11:43 The nonmathematician inside me: c'mon, it's just a typo. The mathematician inside me: Please replace \alpha_n by \frac{1}{\alpha_n}. Plllllleeeeaaaase!!!
if x+y+z=1 and are all positive real numbers, then assuming they are unequal, they will have some strict order x < y < z < 1. This implies that 1/x > 1/y > 1/z > 1, and therefore 1/x+1/y+1/z > 1+1+1, thus 1/x+1/y+1/z > 3. 🎉
What about your algebra series ? Here what I would like to see Cayley Hamilton theorem , rotation matrices, reflection matrices , Gram-Schmidt orthogonalization , QR decomposition, orthogonalization of basis {1,x,..,x^n} with different inner product = \int_{a}^{b} p_{m}(x)q_{n}(x)w(x)dx where [a,b] is interval w(x) is weight function p_{n}(x) is polynomial of degree n q_{m}(x) is polynomial of degree m
What about doing this procedure: Since x+y+z = 1 and x,y,z > 0 then: a) x < x+y+z = 1 => x < 1 => 1 < 1/x b) y < x+y+z = 1 => y < 1 => 1 < 1/y c) z < x+y+z = 1 => z < 1 => 1 < 1/z Then we can use that 3 inequalities to get our result: 1/x + 1/y + 1/z > 1 + 1 + 1 = 3
Very nice way to rigorously put it! It's essentially the "Since x,y and z must sum to 1 and are greater than 0, x,y,z must be less than 1 and hence their reciprocals are greater than 1, and hence the sum of the reciprocals must be greater than 3" logic.
Multivariable calculus. To optimize under constraints, we use Lagrange multipliers. The minimum value of 1/x+1/y+1/z will be found either where ∇(1/x+1/y+1/z) = λ∇(x+y+z-1), or will be found along the boundaries. The boundaries are where x, y, or z is 0. In the limit as either variable approaches positive 0, 1/x+1/y+1/z approaches +∞, so that/those can't be the minimum. ∇(1/x+1/y+1/z) = λ∇(x+y+z-1) = λ, and x+y+z=1 = λ, and x+y+z=1 1/x², 1/y², and 1/z² are all the same. Since they're all positive, x=y=z, meaning x=y=z=⅓. 1/x+1/y+1/z = 9. Therefore, the smallest possible value for 1/x+1/y+1/z under these constraints is 9, which is greater than 3. Thus, 1/x+1/y+1/z>3. Or, alternatively, if we started with all 3 variables at ⅓, then all 3 variables are ⅓. If we tried increasing one or more variables, it would necessarily decrease the other variable(s), thus at least one number is less than ⅓. If we instead tried decreasing one or more variables, the fact remains that at least one variable is less than ⅓. Thus, no matter what, at least one variable is less than or equal to ⅓. The reciprocal of that number is greater than or equal to 3, and since the other two variables are positive, 1/x+1/y+1/x>3. Those who've heard of the pigeonhole principle might find this very familiar. However, if we did it this way, we wouldn't be able to determine the absolute minimum value of 1/x+1/y+1/z.
We know that 1/x + x >= 2. The same idea applies for y and z. So we have: 1/x + x >= 2 (1) 1/y + y >= 2 (2) 1/z + z >= 2 (3) From (1) + (2) + (3) we get that 1/x + 1/y + 1/z >= 5. Since 5 is greater than 3, the problem is solved.
Given x, y, z > 0, s.t. x + y + z = 1. Prove that 1/x + 1/y + 1/z > 3. Without loss of generality assume x >= y >= z. Suppose 1 / x + 1 / y + 1 / z = 1/3. When z = 1/3 => 1 / x + 1 / y + 3 1/3: x >= y > 1/3 x + y > 2/3 But on the other hand x + y + 1/3 < 1 x + y < 2/3 We arrive at a contradiction, so 1/x + 1/y + 1/z > 3
Yeah, I think you got the simplest proof 👍 still the list of means inequalities comes in handy in many cases, especially knowing the equality is reaching only for case where all elements are equals to the mean.
same idea, I would like to hear the point of view of others on it. x+Y+Z=1 => x1 and 1/z>1 => 1/X+1/Y+1/Z > 3 is this method correct? if yes is it similar to one of the three methods in the video?
There is a one line proof for this: x + y + z = 1, so the average of x,y,z is 1/3. Thus at least one of them is less than or equal to 1/3. WLOG let 0< x=3, so the sum 1/x + 1/y + 1/z >= 1/x >= 3.
With x + y + z = 1 and all of x, y, and z being positive, it's safe to assume that all of them are less than 1. If one of them was 1 or greater that would imply at least one of the other numbers being 0 or negative. x, y, and z are all less than 1, so 1/x, 1/y, and 1/z are all greater than 1. Therefore 1/x + 1/y + 1/z > 3.
2 місяці тому
from Morocco thank you for your clear complete proofs ....we can use that t+1/t greater than 2 applied to x y and z...summing then greater than 6 then use x+y+z=1 we took 1 from booth we get stronger than 3 we get GREATER THAN 5....AM I WRONG OR RIGHT
The method I used was a little unusual but I think is valid: Starting from the inequality, multiply both members for xyz => xy+xz+yz> 3xyz. Observing both terms I thought of the Vieta's formulas, so I put x,y and z as zeros of a third grade function f(X)=(X-x)(X-y)(X-z)= X^3-(x+y+z)X^2+(xy+xz+yz)X-xyz= X^3-X^2+(xy+xz+yz)X-xyz [as x+y+z=1]. Now, if the inequality is true, it implies f(X) should be bigger than this other function g(X)=X^3-X^2+3xyzX-xyz. How can it be proven? We need to prove it for values between 0 and 1, because x,y and z are in this specific range. So, I can write g(X)=X^2(X-1)+3xyzX-3xyz+2xyz=X^2(X-1)+3xyz(X-1)+2xyz=(X-1)(X^2+3xyz)+2xyz. I know that f(X=x)=f(X=y)=f(X=z)=0 but g(X) is less than 0 for x,y and z, because (X-1) will have a negative sign and (X^2+3xyz) is a positive number greater than 2xyz so the result of the function g(X) in x,y and z will be negative. Therefore, f(X)>g(X) for all values betweemn 0 and 1 => the first inequality is true => 1/x + 1/y + 1/z > 3
I think you're saying that because f is 0 at x y and z, then showing that g < 0 at x y and z proves the result. If so, while I agree that X-1 is negative and that X^2 + 3xyz is positive and > 3xyz, I don't think it follows that their product is more negative than 2xyz, as needed to establish that g < 0, because X-1 has size < 1. Comparison of f & g is ultimately comparison of xy+yz+xz & 3xyz because all their other terms agree. The former is indeed strictly greater than the latter in the case in which x y ans z are not all equal because of the GM-AM inequality. You may wish to look at my separate comment showing that. But your post interested me as I too was reminded of Vieta's formula.
As may have already been noted, the last term in the denominator of the HM definition on the blackboard is not a reciprocal as are all terms preceding it which seems an oversight.
It is given that: • x,y,z>0 • x+y+z=1 Prove that (1/x)+(1/y)+(1/z)>3 It raises a question that: Does the problem make sense? Reason: x,y,z are positive but their sum is 1. The only possibility is one of them is 1 and the other two are 0. But it must be shown that sum of their inverse is greater than 3, while dividing by 0 is undefined.
None of these methods are the obvious one, which is as follows: If x+y+z=1, then there are 2 possible cases: CASE 1 x or y or z = 1. In this case, the variables not equal to one are both zero. 1/0 = infinity. 2 x Infinity>3. CASE 2 x, y and z are all < 1. In this case, 1/x, 1/y and 1/z are all > 1. Therefore the sum 1/x + 1/y + 1/z > 3. So, 1/x + 1/y + 1/z > 3 in both cases. QED.
actually, we know that (x-1)^2>0. We expand and we rearrange, we obtain x+1/x>2, this is always valid for y and z: y+1/y>2 and z+1/z>2, if we sum these inequalities, we find x+y+z+1/x+1/y+1/z>6 which gives 1+1/x+1/y+1/z>6 which gives 1/x+1/y+1/z>5>3
I solved it with your Method 2 with exactly the same thoughts. If 1 of these is smaller 1/3 than 1 devided by that number will be bigger 3 resulting in a number bigger 3. If all 1/3 result is 9.
we calculate (1/x+1/y+1/z)-3=(yz+xz+xy-3xyz)/xyz since x,y,z are >0 then xyz>0 Then the signe of (1/x+1/y+1/z)-3 depends on (yz+xz+xy)-3xyz (yz+xz+xy)-3xyz=yz-xyz+xz-xyz+xy-xyz (yz+xz+xy)-3xyz=yz(1-x)+xz(1-y)+xy(1-z) x,y,z are >0 and x+y+z=1 we must have 00 (1/x+1/y+1/z) > 3
my method: x+y+z = 1 and all x,y,z are positive implies that all of them are smaller than 1, so all the reciprocals will be greater than 1 and thus the sum of reciprocals are bigger than 3
x,y and z are positives greater than 0 and the least whole number after 0 is 1. But its given that x + y + z = 1. Therefore x,y and z cannot be whole numbers bcoz each of them is greater than 0 and their sum adds up to 1 (x+y+z =1 is only possible if one of the numbers is negative or two out of three numbers are 0 but in this case neither is possible as its given that all are greater than 0 and all are positives) so x y and z are decimals greater than 0 which can be expressed as 1/a, 1/b, 1/c respectively where a,b and c are whole numbers greater than 1 (they must be greater than 1 or else 1/a, 1/b or 1/c will yield numbers greater than 1 or equal to 1 which isnt possible cuz 1/a + 1/b + 1/c =1). We need to prove that 1/x + 1/y + 1/z > 3 i.e a+b+c>3. Now if a,b and c are all greater than 1, the sum of these three must be >3 bcoz when 1 is only added to itself 3 times its equal to 3. so if numbers greater than 1 are added to each other they must be greater than 3
Very interesting👍 im very bad at mathematics but i have tried to resolve it before i can see how you make it and i proved the answer is bigger than 3😂 X+y+z=1 so x=0.3 y=0.3 z=0.4 so x+y+z=1 right? i used calculator so for 1/0.3+1/0.3+1/0.4 i get 9,16 so >3 so simple😏
If any of x, y or 3, as x, y, z > 0. Now as x + y + z = 1, then not all of x, y, z can be < 1/3, as this would mean x + y + z < 1, which contradicts x + y + z = 1. So, we conclude that 1/x + 1/y + 1/z > 3.
From x + y + z =1 it follows that 1 + (y + z)/x = 1/x and 1 + (x + z)/y = 1/y and 1 + (x + y)/z = 1/z, which in sum gives that 1/x + 1/y + 1/z = 3 + (y + z)/x + (x + z)/y + (x + y)/z. Since x, y, z > 0 it follows that 1/x + 1/y + 1/z > 3.
It can be strictly proved (using the derivative) that the sum of 1/x+1/y+1/z cannot be less than 9, and the minimum value is reached at x=y=z=1/3. For all other variable values, the sum is greater than 9. I did it!
I have a doubt, in the last method, if 1/x+1/y+1/z≥9, then how can we say 1/x+1/y+1/z>3 as the second inequality tells that lhs can take any value greater than 3 for example 4. But from the first equation we get minimum value is 9
My brain isn't working today... ...what I tried was pretty stupid. I substituted z = 1-x-y the the equation and got x - x² + y - y² - 4xy + 3x²y + 3xy² > 0. It is easy to show x - x² + y - y² > 0, but that's all I could do.
since for x positive f(x)=1/x is convex we have by jensens inequality, 1/3(1/x + 1/y + 1/z)>=1/((x+y+z)/3) = 3 so 1/x+1/y+1/z >= 9 so we have the stronger fact that this is larger than or equal 9 so definitly more than 3
If x+y+z=1 and x,y,z are all positive then at least one of x,y,z is less than or equal to 1/3 say x, so 1/x is at least 3 and since y,z also positive the sum is definitely greater than 3.
Given that x + y + z = 1, x, y, z > 0 1/x + 1/y + 1/z > 3 (x + y + z)(1/x + 1/y + 1/z) > 3 (x + y)/z + (y + z)/x + (z + x)/y > 0, which is obviously true as x, y, z > 0
We go a little bit further by using the Cauchy Schwarz Inequality: Let x = a^2, y = b^2, z = c^2, then 1/x + 1/y + 1/z = 1/a^2 + 1/b^2 + 1/c^2, (1/a^2 + 1/b^2 + 1/c^2)(a^2 + b^2 + c^2) >= (1 + 1 + 1)^2 -- (Cauchy Schwarz Inequality) this gives that 1/x + 1/y + 1/z >= 9 In fact, with this method, it can be used to prove up to any finite sum of variables as long as their sum is 1. QED Edit: Correction, the variables also have to be positive to use the substitution
If x, y, z are all positive one and add up to 1 they each have to be less than 1 but then their inverses are each bigger than 1 so 1/x + 1/y + 1/z > 1 + 1 + 1 > 3. That shouldn't take more than 20 seconds
Probably prime Newtonw doesn't want to make shorts)
make it 10 seconds, lol
I have confused as hell when I saw the thumbnail in my notifications. So obvious
@@prog8123 Showing other solutions is good.
@@misterj.a91true lol, I was like wait if none of them can be greater than 1 then the inverse is going to be.
Mean Inequality Chain HM
Simply substitute the sum x+y+z for the 1 in the numerator. You get:
x+y+z=1
1/x+1/y+1/z>3
(x+y+z)/x+(x+y+z)/y+(x+y+z)/z>3
Perform term-by-term division:
1+(y+z)/x+1+(x+z)/y+1+(x+y)/z>3
(y+z)/x+(x+z)/y+(x+y)/z>0
Because x, y, z > 0, the inequality is true.
Intuitively, the symmetric case is x=y=z=1/3, which would yield 9 in the second inequality which is larger than 3. Any other combination of x,y,z will require one of them to be less than 1/3 and another to be more than 1/3. The one that's less than 1/3 alone makes the sum of the reciprocals bigger than 3.
Nice, this was your method 2 :D
Madara Uchiha 🔴👁️
Oh cool, some friend showed me problems like that
I didn't see them before but all these 3 methods are things I intuitively guessed in the past, good to know I wasn't completely wrong
I get it.....kinda 3:34 @@thermitty_qxr5276
Yeah, that's how I did it also after considering doing the full partial derivatives to find the extremum. Before picking a piece of paper, I realized that the order 3 symmetry in x, y, z force the extremum to be in x=y=z, then you just check if you got a minimum or maximum, which is also trivial. But I did not know the chain of means inequality, which I will remember because it's a nice formalisation of intuitions I have used time and time again and each time did a quick verification. This will makes things faster and easier in the future 👍
Thank you for another wonderful and blessed video.
Substituting x + y + z for 1 in each term in 1/x + 1/y + 1/z is so cool and produces a great solution almost immediately.
I was familiar with the GM < or = AM inequality but had not come across those involving HM and QM before, so thank you for helping me to never stop learning.
Another proof using just the GM-AM inequality is as follows.
Let L = 1/x + 1/y + 1/z
Case 1: x=y=z
As in your solution, this means x=y=z=1/3 so L=3+3+3=9. Hence L>3.
Case 2: x y and z not all equal
Adding its 3 fractions,
L = S/xyz (1) where S = yz+zx+xy
By the GM-AM inequality,
S/3 > cube root (yz*zx*xy)
The inequality is strict because it is equal only when yz, zx and xy are all equal, which is not so in this case (because yz=zx means x=y and zx=xy means x=z, contradiction)
Hence S > 3 * cube root (x^2*y^2*z^2)
Observe that x, y and z are all > 0 and < 1. Hence x^2 > x^3 and similarly for y and z
Hence S > 3 * cube root (x^3*y^3*z^3)
= 3xyz = 3S/L from (1)
Hence S > 3S/L. And S>0. Hence L>3.
As others have noted, just 0 < x, y and z < 1 is enough to give 1/x, 1/y and 1/Z all > 1 and hence L>3. So the above proof is overkill but still fun for me to play with.
Again, thank you for your lovely videos and i look forward to seeing the next one. God bless you ❤
It is actually quite easy to show that the inequality >= 9 (stronger than >3)…
Since x+y+z=1 it follows that
(1/x)+(1/y)+(1/z) = (x+y+z) *((1/x)+(1/y)+(1/z)).
Expanding this we get
(1/x)+(1/y)+(1/z) = 3 + (x/y) + (y/x) + (x/z) + (z/x) + (y/z) + (z/y).
Now for any a>0, it is easy to show that (1/a)+a >= 2 with equality iff a=1. Applying this to the above with a=x/y, x/z, y/z we get
(1/x)+(1/y)+(1/z) >= 3 + 2 + 2 + 2 = 9, with equality iff all numbers are equal to 1/3 .
Regardless I enjoyed the video since it gave a few different approaches - drives the point home that there isn’t always a unique approach 😊
Very nice, thank you. Equivalent to replacing 1 with x + y + z in each term for the first step but then working on the fractions to establish a lower bound on their sum greater than simply 0.
finally someone with some basic knowledge about inequalities
Cruder but simpler and sufficient for the problem set is to apply a + 1/a > or = 2 to x, y and z directly to give 1/x+1/y+1/z > or = 2+2+2-(x+y+z) = 5 > 3.
@@Jeremy-i1d yes, good thinking - that would work for the problem at hand!
Thanks for sharing your expertise and enthusiasm. Every time I watch your videos I feel smarter.
It's a pleasure to watch your videos man. Keep it up!
Excellent !! I love seeing multiple approaches
the actual sharp inequality is >=9, this is by cauchy
Very nice!
First method is the simplest and more direct one.
Good to know about those terms (i.e. HM, GM and QM)
Thanks again
You've got yourself another subscriber, my friend.
The simplest proof for me: Given that x + y + z = 1 and x, y, z > 0, we know that each of x, y, and z must be less than 1 because they sum up to one. Therefore, 1/x, 1/y, and 1/z are each greater than 1 because x, y, and z are less than one. Consequently, their sum is strictly greater than 3 because each term is more than one and there are three terms.
Trouve en 3secondes!!!!
How complicated can he make this? Each reciprocal is greater than 1.......QED. He clearly shows that he likes algebra, in most cases it is simply not needed.
@@normtyneships194 The algebra in the first step is pretty useful for someone who might not be strong in logic, showing that the sum is 3 + [a bunch of + terms] puts it in very bluntly that it's greater than 3
Excellent presentation - I started by looking at minimizing with constraints, using calculus - but I really prefer all three of your methods!
All three x, y, z are less than 1, therefore all three 1/x, 1/y, 1/z are greater than 1. So, the addition of those reciprocals will always be greater than 3. That’s it.
Another approach:
It is clear that x < 1, y < 1, and z < 1.
x < 1 implies that yz*x < yz (1)
y < 1 implies that xz*y < xz (2)
z < 1 implies that xy*z < xy (3)
Adding (1), (2), and (3) side by side yields:
3xyz < yz+xz+xy
Dividing both sides by xyz (which is positive) yields:
3 < (yz+xz+xy)/xyz = 1/x +1/y + 1/z Q.E.D.
This so interesting and so well presented and explained. I wish I had had this man as my math teacher when I was at school
Sir's smile is so positive and powerful ❤❤
Love this, we live in an age when the entirety of human knowledge is at our fingertips. Every scholarly article is there.
Love your channel!
Hello, nice video. One more method , Since , x>0,y>0,z>0, and x+y+z=1 , so , 0
11:43 The nonmathematician inside me: c'mon, it's just a typo. The mathematician inside me: Please replace \alpha_n by \frac{1}{\alpha_n}. Plllllleeeeaaaase!!!
Method one was the most
direct, but it was good to
see a full discussion. If you
squared x + y + z you could
have used the quadratic mean
also.
Thank you for making these high quality videos, you are helping me stay motivated to keep learning new maths.
Excellent!!! Thanks very much, Sir! ❤❤❤🙏🙏🙏🪷🪷🪷🙇♂️
if x+y+z=1 and are all positive real numbers, then assuming they are unequal, they will have some strict order x < y < z < 1. This implies that 1/x > 1/y > 1/z > 1, and therefore 1/x+1/y+1/z > 1+1+1, thus 1/x+1/y+1/z > 3. 🎉
Set it up as a constrained minimization problem using a Lagrange multiplier; you immediately get x = y = z.
Notice that x+1/x ≥2 for all x>0. The case obviously holds y, z so x+1/x + y+ 1/y z + z+ 1/z ≥6 => 1 + 1/x + 1/y + 1/z ≥ 5 > 3 hence proven
What about your algebra series ?
Here what I would like to see
Cayley Hamilton theorem , rotation matrices, reflection matrices , Gram-Schmidt orthogonalization , QR decomposition,
orthogonalization of basis {1,x,..,x^n} with different inner product = \int_{a}^{b} p_{m}(x)q_{n}(x)w(x)dx
where [a,b] is interval
w(x) is weight function
p_{n}(x) is polynomial of degree n
q_{m}(x) is polynomial of degree m
What about doing this procedure:
Since x+y+z = 1 and x,y,z > 0 then:
a) x < x+y+z = 1 => x < 1 => 1 < 1/x
b) y < x+y+z = 1 => y < 1 => 1 < 1/y
c) z < x+y+z = 1 => z < 1 => 1 < 1/z
Then we can use that 3 inequalities to get our result:
1/x + 1/y + 1/z > 1 + 1 + 1 = 3
Very nice way to rigorously put it! It's essentially the "Since x,y and z must sum to 1 and are greater than 0, x,y,z must be less than 1 and hence their reciprocals are greater than 1, and hence the sum of the reciprocals must be greater than 3" logic.
Multivariable calculus.
To optimize under constraints, we use Lagrange multipliers. The minimum value of 1/x+1/y+1/z will be found either where ∇(1/x+1/y+1/z) = λ∇(x+y+z-1), or will be found along the boundaries. The boundaries are where x, y, or z is 0. In the limit as either variable approaches positive 0, 1/x+1/y+1/z approaches +∞, so that/those can't be the minimum.
∇(1/x+1/y+1/z) = λ∇(x+y+z-1)
= λ, and x+y+z=1
= λ, and x+y+z=1
1/x², 1/y², and 1/z² are all the same. Since they're all positive, x=y=z, meaning x=y=z=⅓. 1/x+1/y+1/z = 9. Therefore, the smallest possible value for 1/x+1/y+1/z under these constraints is 9, which is greater than 3. Thus, 1/x+1/y+1/z>3.
Or, alternatively, if we started with all 3 variables at ⅓, then all 3 variables are ⅓. If we tried increasing one or more variables, it would necessarily decrease the other variable(s), thus at least one number is less than ⅓. If we instead tried decreasing one or more variables, the fact remains that at least one variable is less than ⅓. Thus, no matter what, at least one variable is less than or equal to ⅓. The reciprocal of that number is greater than or equal to 3, and since the other two variables are positive, 1/x+1/y+1/x>3. Those who've heard of the pigeonhole principle might find this very familiar. However, if we did it this way, we wouldn't be able to determine the absolute minimum value of 1/x+1/y+1/z.
I used the same approach but I really appreciated learning the HM AM chain. That was new to me.
We know that 1/x + x >= 2. The same idea applies for y and z. So we have:
1/x + x >= 2 (1)
1/y + y >= 2 (2)
1/z + z >= 2 (3)
From (1) + (2) + (3) we get that 1/x + 1/y + 1/z >= 5. Since 5 is greater than 3, the problem is solved.
Given x, y, z > 0, s.t. x + y + z = 1. Prove that 1/x + 1/y + 1/z > 3.
Without loss of generality assume x >= y >= z.
Suppose 1 / x + 1 / y + 1 / z = 1/3.
When z = 1/3 => 1 / x + 1 / y + 3 1/3:
x >= y > 1/3 x + y > 2/3
But on the other hand x + y + 1/3 < 1 x + y < 2/3
We arrive at a contradiction, so 1/x + 1/y + 1/z > 3
The expression implies x1.
Therefore 1/x + 1/y +1/z > 1+1+1 =3.
Yeah, I think you got the simplest proof 👍
still the list of means inequalities comes in handy in many cases, especially knowing the equality is reaching only for case where all elements are equals to the mean.
Thank you for your videos, they're excellent to learn maths
since x+y+z=1 and x,y,z>0 then it certainly results
x1
adding the three previous inequalities we obtain
1/x+1/y+1/z >3 as was to be demonstrated.
Wiki is a very good source for objective, rigorous fields!
The subject to which I fight always but i love it more than any other subject ❤
Keep going sir
Greetings, Never Stop Teaching!
If x,y,z are not all equal to 1/3 then by the pigeonhole principle one of them is less than 1/3 Let x3 Then 1/x+1/y+1/z>3
Each of x, y, z is less than 1. Hence each reciprocal is greater than 1. Hence the sum of the reciprocals is greater than 3.
same idea, I would like to hear the point of view of others on it.
x+Y+Z=1 => x1 and 1/z>1 => 1/X+1/Y+1/Z > 3
is this method correct? if yes is it similar to one of the three methods in the video?
There is a one line proof for this:
x + y + z = 1, so the average of x,y,z is 1/3. Thus at least one of them is less than or equal to 1/3. WLOG let 0< x=3, so the sum 1/x + 1/y + 1/z >= 1/x >= 3.
With x + y + z = 1 and all of x, y, and z being positive, it's safe to assume that all of them are less than 1. If one of them was 1 or greater that would imply at least one of the other numbers being 0 or negative.
x, y, and z are all less than 1, so 1/x, 1/y, and 1/z are all greater than 1.
Therefore 1/x + 1/y + 1/z > 3.
from Morocco thank you for your clear complete proofs ....we can use that t+1/t greater than 2 applied to x y and z...summing then greater than 6 then use x+y+z=1 we took 1 from booth we get stronger than 3 we get GREATER THAN 5....AM I WRONG OR RIGHT
The simplest method to prove this is : all x,y,z0. So 1/x,1/y,1/z>1 . Adding all 3 we get 1/x+1/y+1/z>3. Does not even require a pen and paper.
You’re a great mathematician .
WLOG assume that 0
The method I used was a little unusual but I think is valid:
Starting from the inequality, multiply both members for xyz => xy+xz+yz> 3xyz.
Observing both terms I thought of the Vieta's formulas, so I put x,y and z as zeros of a third grade function
f(X)=(X-x)(X-y)(X-z)= X^3-(x+y+z)X^2+(xy+xz+yz)X-xyz= X^3-X^2+(xy+xz+yz)X-xyz [as x+y+z=1].
Now, if the inequality is true, it implies f(X) should be bigger than this other function g(X)=X^3-X^2+3xyzX-xyz.
How can it be proven? We need to prove it for values between 0 and 1, because x,y and z are in this specific range.
So, I can write g(X)=X^2(X-1)+3xyzX-3xyz+2xyz=X^2(X-1)+3xyz(X-1)+2xyz=(X-1)(X^2+3xyz)+2xyz.
I know that f(X=x)=f(X=y)=f(X=z)=0 but g(X) is less than 0 for x,y and z, because (X-1) will have a negative sign and (X^2+3xyz) is a positive number greater than 2xyz so the result of the function g(X) in x,y and z will be negative.
Therefore, f(X)>g(X) for all values betweemn 0 and 1 => the first inequality is true => 1/x + 1/y + 1/z > 3
I think you're saying that because f is 0 at x y and z, then showing that g < 0 at x y and z proves the result. If so, while I agree that X-1 is negative and that X^2 + 3xyz is positive and > 3xyz, I don't think it follows that their product is more negative than 2xyz, as needed to establish that g < 0, because X-1 has size < 1.
Comparison of f & g is ultimately comparison of xy+yz+xz & 3xyz because all their other terms agree. The former is indeed strictly greater than the latter in the case in which x y ans z are not all equal because of the GM-AM inequality. You may wish to look at my separate comment showing that. But your post interested me as I too was reminded of Vieta's formula.
I multiplied both sides by 1 in a form of x + y + z, which leads to your method 1, but with a different motivation.
As may have already been noted, the last term in the denominator of the HM definition on the blackboard is not a reciprocal as are all terms preceding it which seems an oversight.
Q: let X=R^3\{O}, the complement of a point O belong to R^3. then X can be partitioned into Euclidean lines.
sir kindly make video on this Q
It is given that: • x,y,z>0
• x+y+z=1
Prove that (1/x)+(1/y)+(1/z)>3
It raises a question that: Does the problem make sense?
Reason: x,y,z are positive but their sum is 1. The only possibility is one of them is 1 and the other two are 0. But it must be shown that sum of their inverse is greater than 3, while dividing by 0 is undefined.
None of the three numbers can be equal or greater than 1, because the other two must be greater than 0.
None of these methods are the obvious one, which is as follows:
If x+y+z=1, then there are 2 possible cases:
CASE 1
x or y or z = 1. In this case, the variables not equal to one are both zero.
1/0 = infinity.
2 x Infinity>3.
CASE 2
x, y and z are all < 1. In this case, 1/x, 1/y and 1/z are all > 1. Therefore the sum 1/x + 1/y + 1/z > 3.
So, 1/x + 1/y + 1/z > 3 in both cases. QED.
actually, we know that (x-1)^2>0. We expand and we rearrange, we obtain x+1/x>2, this is always valid for y and z: y+1/y>2 and z+1/z>2, if we sum these inequalities, we find x+y+z+1/x+1/y+1/z>6 which gives 1+1/x+1/y+1/z>6 which gives 1/x+1/y+1/z>5>3
I solved it with your Method 2 with exactly the same thoughts. If 1 of these is smaller 1/3 than 1 devided by that number will be bigger 3 resulting in a number bigger 3. If all 1/3 result is 9.
By Cauchy-Schwarz inequality,
(1/x * x + 1/y * y + 1/z * z)^2 3
Method 3 and Cauchy-Schwarz inequality are at high-school level
we calculate (1/x+1/y+1/z)-3=(yz+xz+xy-3xyz)/xyz
since x,y,z are >0 then xyz>0
Then the signe of (1/x+1/y+1/z)-3 depends on (yz+xz+xy)-3xyz
(yz+xz+xy)-3xyz=yz-xyz+xz-xyz+xy-xyz
(yz+xz+xy)-3xyz=yz(1-x)+xz(1-y)+xy(1-z)
x,y,z are >0 and x+y+z=1 we must have 00
(1/x+1/y+1/z) > 3
thanks @Prime_Newton, the formula for harmonic mean that was written on the board seems have some typos. (alpha n must be inverse). am I right?
x=y=z=1/3 1/(1/3)+1/(1/3)+1/(1/3)=9 final answer
my method:
x+y+z = 1 and all x,y,z are positive implies that all of them are smaller than 1, so all the reciprocals will be greater than 1 and thus the sum of reciprocals are bigger than 3
You’re a legend, mate.
great video, intriguing!
x,y and z are positives greater than 0 and the least whole number after 0 is 1. But its given that x + y + z = 1. Therefore x,y and z cannot be whole numbers bcoz each of them is greater than 0 and their sum adds up to 1 (x+y+z =1 is only possible if one of the numbers is negative or two out of three numbers are 0 but in this case neither is possible as its given that all are greater than 0 and all are positives) so x y and z are decimals greater than 0 which can be expressed as 1/a, 1/b, 1/c respectively where a,b and c are whole numbers greater than 1 (they must be greater than 1 or else 1/a, 1/b or 1/c will yield numbers greater than 1 or equal to 1 which isnt possible cuz 1/a + 1/b + 1/c =1). We need to prove that 1/x + 1/y + 1/z > 3
i.e a+b+c>3. Now if a,b and c are all greater than 1, the sum of these three must be >3 bcoz when 1 is only added to itself 3 times its equal to 3. so if numbers greater than 1 are added to each other they must be greater than 3
I said x+y+z=1 and x,y,z>0 implies x1. Therefore 1/x+1/y+1/z>3.
Now prove that the sum is greater than or equal to 9.
Solution: use Cauchy-Schwartz inequality:
(x+y+z)(1/x+1/y+1/z)>=(3)²=9
Very interesting👍 im very bad at mathematics but i have tried to resolve it before i can see how you make it and i proved the answer is bigger than 3😂
X+y+z=1 so x=0.3 y=0.3 z=0.4 so x+y+z=1 right? i used calculator so for 1/0.3+1/0.3+1/0.4 i get 9,16 so >3 so simple😏
If any of x, y or 3, as x, y, z > 0.
Now as x + y + z = 1, then not all of x, y, z can be < 1/3, as this would mean x + y + z < 1, which contradicts x + y + z = 1.
So, we conclude that 1/x + 1/y + 1/z > 3.
f := ( 1/x+1/y+1/z ) * 1 = ( 1/x+1/y+1/z ) * (x+y+z) = x/x + y/y + z/z + positive terms > 3
Cool stuff!
From x + y + z =1 it follows that
1 + (y + z)/x = 1/x and 1 + (x + z)/y = 1/y and 1 + (x + y)/z = 1/z, which in sum gives that 1/x + 1/y + 1/z = 3 + (y + z)/x + (x + z)/y + (x + y)/z.
Since x, y, z > 0 it follows that 1/x + 1/y + 1/z > 3.
sir kindly make series on measure theory
If x, y and z are all positive, and x + y + z = 1, then x < 1, y < 1, z < 1, so 1/x > 1, 1/y > 1, 1/z > 1. Adding them up, we get 1/x + 1/y + 1/z > 3
What is the minimum value that the sum of 1/x+ 1/y + 1/z can take? Can it be less than 9 (when x=y=z), and at what x, y, z does this happen?
It can be strictly proved (using the derivative) that the sum of 1/x+1/y+1/z cannot be less than 9, and the minimum value is reached at x=y=z=1/3. For all other variable values, the sum is greater than 9. I did it!
x+y+z=1 ==> 0
I have a doubt, in the last method, if 1/x+1/y+1/z≥9, then how can we say 1/x+1/y+1/z>3 as the second inequality tells that lhs can take any value greater than 3 for example 4. But from the first equation we get minimum value is 9
My brain isn't working today... ...what I tried was pretty stupid. I substituted z = 1-x-y the the equation and got
x - x² + y - y² - 4xy + 3x²y + 3xy² > 0.
It is easy to show x - x² + y - y² > 0, but that's all I could do.
Good question thanks 👍
At 11:44, while writing HM.
THe last term should be 1/alpha n, though writing mistake need to correct.
x,y,z>0 x+y+z=1 prove that 1/x+1/y+1/z>3 Method 1 observe 1/x +1/y+1/z=(x+y+z)/x+(x+y+z)/y+(x+y+z)/z=3+((y+z)/x+(x+z)/y+(x+y)/z)>3
by the titu's lemma we have 1/x + 1/y + 1/z >= (1+1+1)^2 / (x+y+z) = 9/1 = 9, and since 9 > 3, the original inequality has been proven.
Very easily explained
since for x positive f(x)=1/x is convex we have by jensens inequality,
1/3(1/x + 1/y + 1/z)>=1/((x+y+z)/3) = 3 so 1/x+1/y+1/z >= 9 so we have the stronger fact that this is larger than or equal 9 so definitly more than 3
1/3 = (x+y+z)/3 >= 3/(1/x+1/y+1/z), 1/x+1/y+1/z >= 9 > 3
If x+y+z=1 and x,y,z are all positive then at least one of x,y,z is less than or equal to 1/3 say x, so 1/x is at least 3 and since y,z also positive the sum is definitely greater than 3.
Shouldn't, at the harmonic mean, alpha-n be 1 over?
Yes
1/x+1/y+1/z>3
Given that x + y + z = 1, x, y, z > 0
1/x + 1/y + 1/z > 3
(x + y + z)(1/x + 1/y + 1/z) > 3
(x + y)/z + (y + z)/x + (z + x)/y > 0, which is obviously true as x, y, z > 0
We go a little bit further by using the Cauchy Schwarz Inequality:
Let x = a^2, y = b^2, z = c^2,
then 1/x + 1/y + 1/z = 1/a^2 + 1/b^2 + 1/c^2,
(1/a^2 + 1/b^2 + 1/c^2)(a^2 + b^2 + c^2) >= (1 + 1 + 1)^2 -- (Cauchy Schwarz Inequality)
this gives that 1/x + 1/y + 1/z >= 9
In fact, with this method, it can be used to prove up to any finite sum of variables as long as their sum is 1. QED
Edit: Correction, the variables also have to be positive to use the substitution
Nice!
I loved it❤❤❤
Where were you many many years ago when I was doing additional maths . 😂
By titu formula or Cauchy inequality, this quantity is bigger than 9
greater than or equal to 9. Equality holds when x = y = z = 1/3.
@@shmuelzehavi4940 in fact x=y=z is the case when it equal, but basically its bigger
I think it must be true that the sum is more than or equal to 9. (Edit: yeah, he proved it at the end.)
Titu's lemma
i love your shirt
Good Job Prime Newtons😅
It's an 20 second problem using AM-GM inequity 😂😂
why so much ado? since x, y, z > 0, each of them is less than one, means each reciprocal is greater than one, the rest is follow
I believe it can be shown that 1/x+1/y+1/z >= 9.
9>3 solution
Why did you get 9 when you flipped 1/3? I would have expected 3