Prove $x^3+y^3+3xyz>z^3$ for Triangle Sides

[anemone]

Let $x,\,y,\,z$ be the sides of a triangle. Prove that $x^3+y^3+3xyz>z^3$.

 

[kaliprasad]

Let $x,\,y,\,z$ be the sides of a triangle. Prove that $x^3+y^3+3xyz>z^3$.

Spoiler

we have
$x \gt z-y $
cube both sides to get
$x^3 \gt z^3-y^3 - 3yz(z-y)$

or $x^3+y^3 + 3yz(z-y) \gt z^3$

and as $x> z-y$ so $3xyz > 3yz(z-y)$hence $x^3 + y^3 + 3xyz \gt z^3$

 

[anemone]

Spoiler

we have
$x \gt z-y $
cube both sides to get
$x^3 \gt z^3-y^3 - 3yz(z-y)$

or $x^3+y^3 + 3yz(z-y) \gt z^3$

and as $x> z-y$ so $3xyz > 3yz(z-y)$hence $x^3 + y^3 + 3xyz \gt z^3$

Well done, kaliprasad! And thanks for participating!