$x$, $y$ and $z$ are sides of a triangle - prove that $(x+y-z)(x-y+z)(-x+y+z)\leq xyz$

Question

$x$, $y$ and $z$ are all sides of a triangle. Prove that $(x+y-z)(x-y+z)(-x+y+z)\leq xyz$.

Equality occurs when all sides are the same length. I don't know how to prove that the left side is always less or equal though.

Any help is much appreciated!

Answer 1

writing your inequality with the variables $a,b,c$

$(a+b-c)(a-b+c)(-a+b+c)\le abc$ using the Ravi-Substitution

$a=y+z,b=x+z,c=x+y$ so your inequality is equivalent to

$(x+y)(x+z)(y+z)\geq 8xyz$

this AM-GM.