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

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SUMMARY

The inequality $x^3+y^3+3xyz>z^3$ holds true for the sides of a triangle, where $x$, $y$, and $z$ represent the lengths of the triangle's sides. This conclusion is supported by the properties of triangle inequalities and the specific relationships between the sides. The discussion emphasizes the importance of understanding geometric properties in proving such inequalities.

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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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