An Inequality from RMO 2016
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- Опубліковано 20 жов 2024
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To avoid any tedious multiplication: since a, b, c ≥ 0 set a=tan²x, b=tan²y, c=tan²z. Then the given equality turns into sin²x+sin²y+sin²z=1 or cos²z=sin²x+sin²y. Using AM-GM: cos²z ≥ 2sinx∙siny. Similarly cos²y ≥ 2sinx∙sinz and cos²x ≥ 2siny∙sinz. Finally, abc=tan²x∙tan²y∙tan²z = (sin²x∙sin²y∙sin²z)/(cos²x∙cos²y∙cos²z) ≤ (sin²x∙sin²y∙sin²z)/(2siny∙sinz∙2sinx∙sinz∙2sinx∙siny) = 1/8
I wouldn't say this method is any less tedious. The multiplication isn't even that bad tbh. That said, using trig is an interesting appro
This looks tedious! 😮😄
Very good method of proving the inequality ! God bless you !
Beautiful problem with beautiful thinking! I wish the beginning was a bit more creative, but the rest's Que Bella!
Thank you!
I think this way to get the right answer is coincident even though it looks beautiful, because abc and ab, ac, bc are not equivalent in 2abc+ab+ac+bc=1. Since the max value is based on a=b=c=0.5, the coefficient 2 of abc is cancelled. If we add one more element d, a/(1+a) + b/(1+b) + c/(1+c) + d/(1+d) = 1, we will get ab+ac+ad+bc+bd+cd+2abc+2abd+2acd+2bcd+3abcd=1, using the same way will get wrong answer.
By symmetry you can notice immediately there is an extreme value at a=b=c. Thus they're equal to 1/2, and cubed is 1/8 so you need a maximum. You can let a = 0 to see it's a max and you're done.
For me the immediate thinking is: The sum condition is hard, the inequality is easy, let's change that by setting x = a/(a+1), y, z, ...
Then x+y+z = 1. Prove that (from ax + x = a => a = x/(1-x)) 8xyz less than (1-x)(1-y)(1-z) which is (y+z)(z+x)(x+y) which follows from 3 AM-GMs.
Could you please elaborate last line more
It can be solved by discriminant of quadratic equation i.e. D=b^2-4ac.
Let t=abc and substitute c by t/ab.
Then the given equation is turned into simpler one: t=(1-ab)/(1/a+1/b+2)
If k=1/a+1/b, then ab=a²/(ka-1). Using discriminant, we get ab ≥ 4/k² and the equality holds when a=b=2/k.
So, put a=b=2/k to get t=(k-2)/k². Again, using discriminant, we get t≤1/8.
RMO is the second stage of IMO team selection in India.
The first stage is PRMO. The third one is INMO where the real difficulty is seen. So wait for it, prolly their questions also gonna come on this channel!!
Already there is on Func Eq problem
Dont know nothing about a math olympiads.. Is there any age restriction? is it for school or undergrates only? could an engineer graduate(first to come in mind) at least solve 1 exercise, with or without preparing/practising?
@@federicohansen4561 It is for high school students
In India it is grade 8-12
(Other countries may not allow 8-9 graders in to IMO but they do go through selection stages)
The ultimate aim is to select 6 students to represent respective country at IMO that takes place annually.
Hard without preparation ) :
@@abhinavbhutada9b484 And especially hard without copying, cheating as many do
If a student has studied the conditional extremum of a function of several variables, then
he should use the Lagrange multiplier method to solve the conditional extremum problem.
f(a,b,c)=abc, coupling equation φ(a,b,c)=a/(1+a) + b/(1+b)+c/(1+c)-1 =0.
F(a,b,c)=f(a,b,c)-λ *φ(a,b,c).
It is necessary to solve a system of equations
[∂F/∂a =0, ∂F/∂b=0, ∂F/∂c=0, φ(a,b,c)=0] =>
[bc-λ /(a+1)^2=0, ac-λ /(b+1)^2=0, ab-λ /(c+1)^2=0, a/(1+a) + b/(1+b)+c/(1+c)-1 =0]
We'll have to work a little before we get:
a=b=c=1/2, λ =9/16.
So maxf(a,b,c)= f(1/2,1/2,1/2)=1/8.
It should be max f(a,b,c) = 1/8. Also, don't you need to include constraints a>0, b>0, c>0 in your Lagrangian formulation?
More simple, consider b and c fixed, and let ∆ the value for a for which 'abc' is maximum considering the given relation (a/(1+a) ...). As a, b, c play symetric roles in this equation and in the expression to maximize (abc), if ∆ is a maximum value for a with fixed b and c, so b=∆ too if a and c are considered as fixed, and so c=∆ if a and b are considered as fixed. So abc will be maximal for a=b=c=∆, and thus abc=∆³. Let's calculate ∆. We have : ∆/(1+∆) + ∆/(1+∆) + ∆/(1+∆)=1; 3∆=1+∆; 2∆=1; ∆=1/2 and the maximum value for abc is thus ∆³=(1/2)³=1/8.Of course it's a maximum because for instance, for a=b=3/4, (3/4)(2*4/7)+c/(1+c)=1; 1-6/7=c/(1+c);1/7=c/(1+c);1+c=7c;c=1/6; and abc=(3/4)*(3/4)*(1/6)=9/16*1/6=3/32=0.0,9375 < 1/8 (=0,125), so 1/8 is a maximum. And you can do the same thing for relation more complex if they are just symetric. But it's "meta-math", a kind of, but it's also math ;)
This is like a hard-ish question from a standardised test in Vietnam for middle school student to get accepted to a decent high school lol. Yes, you heard me right, standardised test, which mean everyone has to take it. Kinda equavilent to the SAT in the US, but we are supposed to do this question in less than 20 minutes. Anyway this is my solution:
a/(1+a) + b/(1+b) + c/(1+c) = 1
b/(1+b) + c/(1+c) = 1 - a/(1+a) = 1/(1+a)
Using the AM-GM Inequality (This is very easy to prove) for the [b/(1+b) + c/(1+c)] expression, we have:
b/(1+b) + c/(1+c) >= 2 . sqrt(bc/(1+b)(1+c)). The equality only happen when 1/(1+b) = 1/(1+c)
Therefore 1/(1+a) >= 2 . sqrt(bc/(1+b)(1+c))
Similarly: 1/(1+b) >= 2 . sqrt(ac/(1+a)(1+c)). The equality only happen when 1/(1+a) = 1/(1+c)
1/(1+c) >= 2 . sqrt(ab/(1+a)(1+b)). The equality only happen when 1/(1+a) = 1/(1+b)
Multiple these three expression together, we have:
1/(1+a)(1+b)(1+c) >= 8 . sqrt(b.c.a.c.a.b/(1+b)(1+c)(1+a)(1+c)/(1+a)(1+b))
1/(1+a)(1+b)(1+c) >= 8 . abc / (1+a)(1+b)(1+c)
Multiply (1+a)(1+b)(1+c) for both side, we have 8 . abc
Thanks for the amazing videos ❤❤
Glad you like them! Thanks for the kind words! 🥰
Elegantly presented
Glad you think so!
@@SyberMath I don't see why anyone would ever think of the AM GM inequality so why do it that way? Surely there's another more intuitive and maybe logically clear way?
@@leif1075 bruh AM-GM has to be one of the most elemental inequality in math. Like bro, this inequality is trivial to prove (for 2 numbers). Middle school students in Vietnam when looking at an inequality question like this will immediately think of the AM-GM inequality. Like, you can only get so far when you just do pure algebra malnipulation.
@@kienthanhle6230 I believe you but my point is could thise students or Ramanujan or me or youanybody think [f the inequality if they hadn't been taught it or pointed it out..I don't see anyone organically intuitively or nonintuitively but naturally logically deriving this..do you? Would you?
@@SyberMath So isn't there another way to do this without AM fm??
Nice
Thank you!
Thank you
You're welcome
Love your Videos!
very nice Solution
Thank you very much! 😊
what if we suppose that abc > 1/8 and while the equation (... = 1) and find a solution for the equation as follow ( first term = 0 {a/1+a = 0} and second and 3 term = 1) and find that bc=1 and we know that a = 0 so abc = 0 which is absurd so abc
i think i just made i mistake supposing that abc> 1/8 means a != 0 and b != 0 c != 0 while in the equation is true (..... = 1) but there is a solution where a/1+a = 0 and the 2nd and 3 term = 1 which is absurd because we gonna find a = 0 so abc
Huh that was straight forward although long. I enjoy this style haha very similar to how I approach stuff
Can please more explain i have slowly understand
perfect⭐
Thank you! 🥰
QUESTION :
Sybermath showed abc=1/8 if a=b=c=1/2. But can 1/8 be attained if a,b,c are NOT all equal??
Good question! I think it can be.
Set a=8/bc and sub
www.wolframalpha.com/input?i=b%2F%281%2Bb%29%2Bc%2F%281%2Bc%29%3D%28bc%29%2F%28bc%2B8%29
Any ideas? Suggestions?
Nah, the abc = 1/8 hold true if and only if 1/1+a = 1/1+b = 1/1+c a = b = c (=1/2). You can check out my solution for this at the comment section to see how did I derive that
Nice solution
Thanks!
great
bài bất đẳng thức phân thức hay quá.
Bạn có thể vẫn dùng giả thuyết để đánh giá a + b + c.
A, b, and c is equal to 1/3
No, *a,* b, and c *are* each equal to *1/2* if the sum is to be 1 and the product is to be 1/8.
Alright, here is the second method, it requires no simplification from the equality!
Let a=x/(y+z) ; b=y/(z+x) ; c=z/(x+y) (for x,y,z > 0)
This way, we won't have to worry about the condition of x,y,z from ab + bc + ca + 2abc = 1, except for x,y,z > 0
(because if you substitute those into the original equality, it will be 1=1, which is true for all x,y,z)
Hence, we only need to prove (xyz)/[(x+y)(y+z)(z+x)] ≤ 1/8
This time, we will use AM-GM:
x+y ≥ 2√xy
y+z ≥ 2√yz
z+x ≥ 2√zx
Multiply all three inequalities together:
(x+y)(y+z)(z+x) ≥ 8√xy√yz√zx = 8xyz
[(x+y)(y+z)(z+x)]/(xyz) ≥ 8
Hence, (xyz)/[(x+y)(y+z)(z+x)] ≤ 1/8
Therefore, abc ≤ 1/8
The equality is only attained if x=y=z a=b=c=1/2
i have fixed the solution, simplification of a/(a+1)+b/(b+1)+c/(c+1)=1 is not necessary.
Great method! With the same method, we may add more elements? if we have 4, a b c d, abcd ≤ 1/81 and 5 a b c d e, abcde ≤ 1/1024
This is super confusing..does anyone actually understand jt..why x y and z..andbifbthey all equal 1..why not one variable??
@@gdtargetvn2418 What do you mean??
There MUST BE. Away to solve without the AM GM inequality nonsense since a lot of ppl would never think of that...
The question does not say that a, b c, are positive, how can you use the AMGM inequality here?
At the beginning of the video, I say that!
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