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

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

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 $x^3 + y^3 + 3xyz \gt z^3$
 
kaliprasad said:
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!
 
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