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Prove $(-a+b+c)(a-b+c)(a+b-c) \leq abc$, where $a, b$ and $c$ are positive real numbers

The title says it all. I've been trying to prove this for hours! Help me!

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marked as duplicate by Norbert, Berci, froggie, Thomas, Cameron Buie Nov 13 '12 at 23:22

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

try expanding the expression and apply triangle inequality – user31280 Nov 13 '12 at 21:40
when I said that "I've been trying to prove this for hours", what I meant was I've been trying to incorporate triangle inequality in order to prove this. The thing is that both abc and (a+b−c)(b+c−a)(c+a−b) are greater or equal than 0. And I can't seem to connect them together in one equation... – Dekac Nov 13 '12 at 21:45
up vote 3 down vote accepted

What I did was wrong.

Use the substitution $a=x+y$, $b=y+z$ and $c=z+x$ in the initial inequality to have $$8xyz\le(x+y)(y+z)(z+x)$$ and that follows directly from $AM-GM$ since

$$\begin{array}{ll}x+y\ge 2\sqrt{xy}\\ y+z\ge 2\sqrt{yz}\\ z+x\ge 2\sqrt{zx}\\(x+y)(y+z)(z+x)\ge 8\sqrt{x^2y^2z^2}=8xyz\\ \end{array}$$

NB: $AM-GM$ applies only because the sides of a triangle are positive, thus $a,b,c>0$.

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something is wrong, the inequality says $\le$ and i have negative values in my expression. – user31280 Nov 13 '12 at 21:55
I think you screwed up the multiplication. – Dekac Nov 13 '12 at 21:56
nope, your multiplication is good. – Dekac Nov 13 '12 at 22:01

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