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?
 A: 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.
A: HINT: To do it without linear algebra, expand both sides and subtract like terms to leave
$$a^2y^2+a^2z^2+b^2x^2+b^2z^2+c^2x^2+c^2y^2=2abxy+2acxz+2bcyz\;.$$
Notice that you can rearrange this as
$$(a^2y^2-2abxy+b^2x^2)+(a^2z^2-2acxz+c^2x^2)+(b^2z^2-2bcyz+c^2y^2)=0\;,$$
or
$$(ay-bx)^2+(az-cx)^2+(bz-cy)^2=0\;.$$


*

*What can you conclude about $ay-bx$, $az-cx$, and $bz-cy$?  

*What can you conclude about $a(x+y+z)$ and $x(a+b+c)$?

A: $$a^2x^2+a^2y^2+a^2z^2+b^2x^2+b^2y^2+b^2z^2+c^2x^2+c^2y^2+c^2z^2=a^2x^2+b^2y^2+c^2z^2+2axby+2axcz+2byxz$$
$$a^2y^2+a^2z^2+b^2x^2+b^2z^2+c^2x^2+c^2y^2=2axby+2axcz+2byxz$$
$$a^2y^2-2axby+b^2x^2+a^2z^2-2axcz+c^2x^2+b^2z^2-2byxz+c^2y^2=0$$
$$(ay-bx)^2+(az-cx)^2+(bz-cy)^2=0$$
From here the answer is clear. All three terms are zero.
