<s>[INST] Help with inequality please 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. 
 [/INST] Given $\quad(a+b)(b+c)(c+a)  \ge   8(a+b-c)(b+c-a)(c+a-b)\quad$ show $x\ge y$
$$x=(a+b)(b+c)(c+a) \quad=a^2 b + a^2 c + a b^2 + 2 a b c + a c^2 + b^2 c + b c^2$$
$$y=8(a+b-c)(b+c-a)(c+a-b) \\ =
-8 a^3 + 8 a^2 b + 8 a^2 c + 8 a b^2 - 16 a b c + 8 a c^2 - 8 b^3 + 8 b^2 c + 8 b c^2 - 8 c^3$$
We negate the RHS and subtract it from both sides by addition so $x-y \ge 0$
$$x-y\quad =\quad a^2 b + a^2 c + a b^2 + 2 a b c + a c^2 + b^2 c + b c^2\\
+8 a^3 - 8 a^2 b - 8 a^2 c - 8 a b^2 + 16 a b c - 8 a c^2 + 8 b^3 - 8 b^2 c - 8 b c^2 + 8 c^3\\
=8a^3-7a^2b-7a^2c-7ab^2+18abc-7ac^2+8b^3-7b^2c-7bc^2+8c^3\quad \ge\quad 0\\
\implies 8(a^3+b^3+c^3)\quad\ge\quad 7(a^2b+a^2c+ab^2+ac^2+b^2c+bc^2)$$
If $\quad a=b=c=1\quad$ then $\quad 24\ge 42\quad $which is a contradiction.
If $\quad a=b=c=2\quad$ then $\quad 192 \ge 336\quad $ which is a contradiction.
If $\quad a=b=c=3\quad$ then $\quad 648 \ge 1134\quad $ which is a contradiction.
This math could be wrong but it appears that the original statement is reversed.
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