Find Maximum $(xyz)^2$ Given $x+y+z=0, x^2+y^2+z^2=2015$

[anemone]

If $x,\,y,\,z$ are three real numbers such that $x+y+z=0$ and $x^2+y^2+z^2=2015$, find the maximum of $(xyz)^2$.

 

[Albert1]

If $x,\,y,\,z$ are three real numbers such that $x+y+z=0---(1)$ and $x^2+y^2+z^2=2015---(2)$, find the maximum of $(xyz)^2$.

Spoiler

let $xyz=k$
from $(1)(2):xy+yz+zx=\dfrac {-2015}{2}$
$x,y,z$ are three roots of:$t^3-\dfrac{2015 t}{2}-k=0$
$k=t^3-\dfrac {2015t}{2}$
$max(k)$ occurs when $t^2=\dfrac{2015}{6}$
$\therefore k=t(t^2-\dfrac {2015}{2})=t(\dfrac {-2015}{3})$
and $max(xyz)^2=max(k^2)=\dfrac{2015^3}{54}$

 

[anemone]

Thanks for participating, Albert!:)

Your answer is correct, of course!