If $a+b+c=0$ prove that $ \frac{a^2+b^2+c^2}{2} \times \frac{a^3+b^3+c^3}{3} = \frac{a^5+b^5+c^5}{5} $ 
If $a+b+c=0$, for $a,b,c \in\mathbb R$, prove
$$ \frac{a^2+b^2+c^2}{2} \times \frac{a^3+b^3+c^3}{3} = \frac{a^5+b^5+c^5}{5} $$

I've tried squaring, cubing, etc. the $a+b+c=0$, but I've just dug myself in.
Is there a more elegant way to prove this, or a way to find the right "trick"?

$M= (a^2 + b^2 + c^2)(a^3 + b^3+c^3)=$
$a^5 + b^5 + c^5 + a^3b^2 +a^3c^2 +a^2b^3 + b^3c^2 + a^2c^3 +b^2c^3$
$= (a^5 + b^5 + c^5) + L$ where
$L = a^3b^2 +a^3c^2 +a^2b^3 + b^3c^2 + a^2c^3 +b^2c^3$
$= a(a^2b^2 + a^2c^2) + b(a^2b^2 + b^2c^2) + c(a^2c^2 + b^2b^2)$
$= (a+b+c)(a^2b^2 + a^2c^2 + b^2c^2) - ab^2c^2 -a^2bc^2 - a^2b^2c$
$= - ab^2c^2 -a^2bc^2 - a^2b^2c= -abc(ab + ac + bc)$
On the other hand:
$a^2 + b^2 + c^2 = (a+b+c)^2 - 2(ab+ac+bc)=-2(ab+ac + bc)$
$= \frac {2L}{abc}$
while $0=(a+b+c)^3 = (a+b)^3 + 3(a+b)^2c + 3(a+b)c^2 + c^3$
$= a^3 + 3a^2b + 3ab^2 + b^3 +3a^2c + 6abc + 3b^2c + 3ac^2 + 3bc^2 + c^3$
$= a^3 + b^3 + c^3 + 3a(ab + ac) +3b(ab + bc) + 3c(ac+ bc) + 6abc$
$=a^3+b^3+c^3 +3(a+b+c)(ab + ac + bc) - 9abc+6abc$
$=a^3 + b^3 +c^3 -3abc$
So $a^3 + b^3+c^3 = 3abc$.
So $M =(a^2+b^2 + c^2)(a^3 + b^3 + c^3) = \frac {2L}{abc}*3abc = 6L$
So $M = a^5 + b^5 + c^5 + L$ and $M = 6L$ so $5L=a^5 + b^5 +c^5$.
And we have:
$\frac {a^2+b^2+c^2}2\times\frac {a^3+b^3 + c^3}3 = \frac M6 = L = \frac{a^5+b^5 + c^5}5$
Yes.  Long and tedious and little insight.  But methodical.
