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Monday, November 30, 2009

Solution: 3 Variable Inequality

Original problem: http://dharmath.thehendrata.com/2009/11/24/3-variable-inequality/

For $a,b,c > 0$, prove that:

$! (ab(a+b) + bc(b+c) + ca(c+a))^2 \geq 4abc(a+b+c)(a^2+b^2+c^2)$

First Solution


WLOG, we assume $a \geq b \geq c$.

From AM-GM: $RHS \leq (ac(a+b+c) + b(a^2+b^2+c^2))^2$

But $ac(a+b+c) + b(a^2+b^2+c^2) \leq ab(a+b) + bc(b+c) + ca(c+a) \iff b(b-a)(b-c) \geq 0$

Second Solution


We will use the usual notation $S(k,l,m)$ to denote the symmetric sum: $S(k,l,m) = \sum a^kb^lc^m$ where the sum is over all permutations of $a,b,c$.

$ab(a+b)+bc(b+c)+ca(c+a) = \sum a^2b = S(2,1,0)$.

$4(a+b+c)(a^2+b^2+c^2) = S(1,0,0).S(2,0,0) = 2S(3,0,0) + 4S(2,1,0)$

So $RHS = abc(2S(3,0,0) + 4S(2,1,0)) = 2S(4,1,1) + 4S(3,2,1)$

And $LHS = S(2,1,0).S(2,1,0) = S(4,2,0) + S(3,3,0) + S(4,1,1)+ 2S(3,2,1) + S(2,2,2)$

So we need to prove:

$! S(4,2,0) + S(3,3,0) + S(2,2,2) \geq S(4,1,1) + 2S(3,2,1)$

By Muirhead, $S(4,2,0) \geq S(4,1,1)$, so we only need to prove

$! S(3,3,0) + S(2,2,2) \geq 2S(3,2,1)$

$! \iff S(3,3,3) (S(0,0,-3) + S(-1,-1,-1)) \geq 2S(3,3,3) S(0,-1,-2)$

$! \iff S(0,0,-3) + S(-1,-1,-1) \geq 2S(0,-1,-2)$

which is true by Schur

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