# Let a, b, and c be the sides of a triangle. Prove that $(a+b-c)(b+c-a)(c+a-b)\leq abc$ [duplicate]

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

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## marked as duplicate by userNaN, Berci, froggie, Thomas, Cameron BuieNov 13 '12 at 23:22

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

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 $a,b,c>0$.
something is wrong, the inequality says $\le$ and i have negative values in my expression. –  user31280 Nov 13 '12 at 21:55