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. 
 A: 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? 
A: 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 $a,b\ge 0$.
A: Download the solution of the 2008 question paper from the SA Mathemataics Foundation website here - http://www.samf.ac.za/QuestionPapers.aspx
A: Since our inequality is symmetric, it's enough to assume that $a\geq b\geq c$, which gives:
$$\prod_{cyc}(a+b)-8\prod_{cyc}(a+b-c)=$$
$$=\sum_{cyc}\left(a^2b+a^2c+\frac{2}{3}abc\right)-8\sum_{cyc}\left(-a^3+a^2b+a^2c-\frac{2}{3}abc\right)=$$
$$=\sum_{cyc}(8a^3-7a^2b-7a^2c+6abc)=$$
$$=\sum_{cyc}(8a^3-4a^2b-4a^2c-(3a^2b+3a^2c-6abc))=$$
$$=\sum_{cyc}4(a-b)^2(a+b)-3\sum_{cyc}c(a-b)^2=\sum_{cyc}(a-b)^2(4a+4b-3c)\geq$$
$$\geq(a-c)^2(4a+4c-3b)+(b-c)^2(4b+4c-3a)\geq$$
$$\geq(b-c)^2(4a+4c-3b)+(b-c)^2(4b+4c-3a)=(b-c)^2(a+b+8c)\geq0.$$
A: 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.
