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?

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
$$(ay-bx)^2+(az-cx)^2+(bz-cy)^2=0\;.$$


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

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

