<s>[INST] Prove that $a(x+y+z) = x(a+b+c)$ If $(a^2+b^2 +c^2)(x^2+y^2 +z^2) = (ax+by+cz)^2$
Then prove that $a(x+y+z) = x(a+b+c)$
I did expansion on both sides and got:
$a^2y^2+a^2z^2+b^2x^2+b^2z^2+c^2x^2+c^2y^2=2(abxy+bcyz+cazx) $ 
but can't see any way to prove $a(x+y+z) = x(a+b+c)$. How should I proceed?
 [/INST] By C-S inequality, $(ax+by+cz)^2\le (a^2+b^2+c^2)(x^2+y^2+z^2)$ with equality iff $(x,y,z)=\lambda(a,b,c)$ for some $\lambda$ or $(a,b,c)=(0,0,0)$. But, if $(a,b,c)=(0,0,0)$, the problem is trivially true. If it not the case, then  $x=\lambda a$, $y=\lambda b$ and $z=\lambda c$.
Then $x+y+z=\lambda(a+b+c)$. Multiplying by $a$ both sides and remembering $x=\lambda a$ yield us the proof.
</s>