2014/065) if x,y,z are sides of a triangle prove that

$x^3+ y^3 + 3xyz \gt z^3$
proof:
we have as x,y,z are sides of triangle $x \gt z- y$
cube both sides to get
$x^3 \gt z^3-y^3-3yz(z-y)$
or
$x^3+y^3 + 3yz(z-y) \gt z^3$
as $z - y \lt x$ we get
$x^3+y^3+3xyz \gt z^3$