Multiplying by abc on 3=1/a + 1/b + 1/c implies ab+ac+bc=3. expanding (a-1)(b-1)(c-1) and subbing in the different values we know shows (a-1)(b-1)(c-1)=0 so at least one of a,b,c=0. wlog we say a=1. b=1/c so sub into b+c=2 and find (c-1)^2=0 so c=1 and hence b=1. thus (a,b,c)=(1,1,1) is the only sol.
my pure algebra approach- I hadn't heard of any of the formulae or techniques mentioned in the video so here is how I solved it. ab+bc+ac=3 ac+bc+ab=3 (1.) (rearranged) a+b+c=3 multiply by c ac+bc+c^2=3c (2.) c^2-ab=3c-3 (2. -1.) abc=1, ab=1/c c^2-1/c=3c-3 (multiply by c) c^3-1=3c^2-3c c^3-3c^2+3c-1=0 (c-1)^3=0 so c=1 only then c=1 so ab=1, using a+b+c=3 a+b=2 (3.) a^2+2ab+b^2=4 a^2+b^2=2 1/a+1/b=2 (multiply by ab) b+a=2ab so by (3.), a+b=b+a, so 2ab=2, b=1/a sub 1/a for b in a^2+b^2=2 a^2-2+1/a^2=0 a^4-2a^2+1=0 solve for (a^2-1)^2=0, a=+-1 but +ve so =1 so b=1 also therefore a,b,c=1,1,1
Given a , b , c > 0 a + b + c = 3 1/a + 1/b + 1/c = 3 ab + bc + ca = 3 abc = 1 let a,b,c be the roots of the polynomial f(x) = x³ - 3x² + 3x - 1 = 0 f'(x) = 3x² - 6x + 4 > 0 for all x € R Then f(x) is increasing function for all x > 0 => a = b = c Therefore , a = b = c = 1
Do you need f(x) or can we start that part by just considering (x-a)(x-b)(x-c)? Also, the part at 4:56 seemed a little hand-wavy to me. Maybe you could show a proof of that in another video.
whats hand-wavy about it, if we define a polynomial with roots a, b and c, and we work out that afterwards the polynomials roots are purely 1, we know a,b and c are all 1?
I was being a bit hand-wavy to be fair. To clarify, we show that f(x) is a cubic with roots 1, 1 and 1, and since there is 'no more room' for any other roots (as f is a cubic), it must follow that the roots coincide with those as defined in f, namely a, b and c.
Multiplying by abc on 3=1/a + 1/b + 1/c implies ab+ac+bc=3.
expanding (a-1)(b-1)(c-1) and subbing in the different values we know shows (a-1)(b-1)(c-1)=0 so at least one of a,b,c=0. wlog we say a=1. b=1/c so sub into b+c=2 and find (c-1)^2=0 so c=1 and hence b=1. thus (a,b,c)=(1,1,1) is the only sol.
Nice solution! Thanks for sharing!
my pure algebra approach- I hadn't heard of any of the formulae or techniques mentioned in the video so here is how I solved it.
ab+bc+ac=3
ac+bc+ab=3 (1.) (rearranged)
a+b+c=3
multiply by c
ac+bc+c^2=3c (2.)
c^2-ab=3c-3 (2. -1.)
abc=1, ab=1/c
c^2-1/c=3c-3 (multiply by c)
c^3-1=3c^2-3c
c^3-3c^2+3c-1=0
(c-1)^3=0 so c=1 only
then c=1 so ab=1, using a+b+c=3
a+b=2 (3.)
a^2+2ab+b^2=4
a^2+b^2=2
1/a+1/b=2 (multiply by ab)
b+a=2ab
so by (3.), a+b=b+a, so 2ab=2,
b=1/a
sub 1/a for b in a^2+b^2=2
a^2-2+1/a^2=0
a^4-2a^2+1=0
solve for (a^2-1)^2=0, a=+-1 but +ve so =1
so b=1 also therefore a,b,c=1,1,1
Given a , b , c > 0
a + b + c = 3
1/a + 1/b + 1/c = 3
ab + bc + ca = 3
abc = 1
let a,b,c be the roots of the polynomial
f(x) = x³ - 3x² + 3x - 1 = 0
f'(x) = 3x² - 6x + 4 > 0 for all x € R
Then f(x) is increasing function for all x > 0
=> a = b = c
Therefore , a = b = c = 1
@@nonroutineganit7682 ah I like this approach, I hadn't thought of it this way!
What is the blackboard software and pen you use to present online ?
@@Mo-zj6hz I use bitpaper at the moment, but I'm sure there may be better ones! And I just use a basic writing tablet
@JPiMaths thanks and to record your screen ?
Do you need f(x) or can we start that part by just considering (x-a)(x-b)(x-c)? Also, the part at 4:56 seemed a little hand-wavy to me. Maybe you could show a proof of that in another video.
whats hand-wavy about it, if we define a polynomial with roots a, b and c, and we work out that afterwards the polynomials roots are purely 1, we know a,b and c are all 1?
@ because polynomials factor uniquely just like natural numbers do?
I was being a bit hand-wavy to be fair. To clarify, we show that f(x) is a cubic with roots 1, 1 and 1, and since there is 'no more room' for any other roots (as f is a cubic), it must follow that the roots coincide with those as defined in f, namely a, b and c.
Can't you just use 1/n < 1 for all n>1 and 1/n = 1 only for n=1 therefore all the numbers must be 1 else 1/a + 1/b + 1/c < 3
No because what about 1/n when n
@artemis477 for positive integers?
Nice try, but n needn't be an integer.
Oh yeah lol helps if I read the question
@@JPiMathsdoesn't matter if n is an integer what matters is n>1 in order for 1/n