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Once again I have come across an olympiad-type problem which probably requires some sort of insight even though it looks simple. The question is as follows:

Let $a$, $b$ and $c$ be positive real numbers. Prove that:

$(a+b)(b+c)(c+a)$ $\geqslant$ $8(a+b-c)(b+c-a)(c+a-b)$

I have tried to multiply out the LHS but unfortunately it didn't get me much...

I found that if one of $a$, $b$ or $c$ is greater than or equal to the sum of the other two, then the inequality is trivially true, since LHS is positive while RHS isn't.

Would there be a quick and easy formula or known inequality that I could use to make this problem simpler? Or is this just a 'bash-and-solve' type question?

Any help, comments or edits are greatly appreciated! Thanks! :)

This question appeared in the South African Mathematics Olympiad in 2008.

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Hint: We may make the assumption that each of the terms on the RHS are positive.

Hint: Use the substitution $$ x = a+b - c \\ y = b+c -a \\ z = c+a - b \\$$

What happens now?

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By homogeneity, we may assume wlog that $a + b + c = 1$. We want to minimize $f(a,b,c) = \left( a+b \right) \left( b+c \right) \left( c+a \right) -8\, \left( a+b-c \right) \left( b+c-a \right) \left( c+a-b \right)$ on the triangle $a+b+c=1$, $a,b,c\ge 0$. Critical points with $a,b,c>0$ are found using a Lagrange multiplier: I get $(1/3,1/3,1/3)$ with $f(1/3,1/3,1/3) = 0$ and $(31/63, 31/63, 1/63)$ and its permutations with $f(31/63, 31/63, 1/63) = 1000/3969$. We must also look at the boundary, but I find that $f(a,b,0) = (a+b)(8 a^2 - 15 a b + 8 b^2) \ge 0$ for $a,b\ge 0$.

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:hi .why do you assume a + b + c = 1? – Maisam Hedyelloo Mar 19 '13 at 21:25
@MaisamHedyelloo if not, divide all of $\{a,b,c\}$ by the sum and it cancels out from the original inequality – gt6989b Mar 19 '13 at 21:43
Because all terms are homogeneous of degree $3$, i.e. change $a,b,c$ to $ta, tb, tc$ and you multiply everything by $t^3$. – Robert Israel Mar 19 '13 at 22:50

Download the solution of the 2008 question paper from the SA Mathemataics Foundation website here -

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Oh thanks, when I tried to access the answers previously the page didn't load for some reason, but now it does :| Thanks! :) – mathsnoob Mar 26 '13 at 6:32

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