<s>[INST] Prove that if $\frac{x+y}{3a-b}=\frac{y+z}{3b-c}=\frac{z+x}{3c-a}$ then $\frac{x+y+z}{a+b+c}=\frac{ax+by+cz}{a^2+b^2+c^2}$ If $\displaystyle\frac{x+y}{3a-b}=\frac{y+z}{3b-c}=\frac{z+x}{3c-a}$ then prove that $\displaystyle\frac{x+y+z}{a+b+c}=\frac{ax+by+cz}{a^2+b^2+c^2}$
I tried to prove this in many ways. First, I tried to multiply the first equality by $\displaystyle(3a-b)(3b-c)(3c-a)$, but then it seems too complicated. Is there any easy method?
 [/INST] If we set
$$k=\frac{x+y}{3a-b}=\frac{y+z}{3b-c}=\frac{z+x}{3c-a},$$
then we have
$$x+y=(3a-b)k,\ \ y+z=(3b-c)k,\ \ z+x=(3c-a)k.$$
Hence, adding these three gives us
$$2(x+y+z)=(3a-a+3b-b+3c-c)k\iff \frac{x+y+z}{a+b+c}=k\tag 1$$
On the other hand, we have
$$a(x+y)=a(3a-b)k,\ \ b(y+z)=b(3b-c)k,\ \ c(z+x)=c(3c-a)k$$
$$\Rightarrow (ax+by+cz)+(ay+bz+cx)=\{3(a^2+b^2+c^2)-(ab+bc+ca)\}k\tag2$$
$$b(x+y)=b(3a-b)k,\ \ c(y+z)=c(3b-c)k,\ \ a(z+x)=a(3c-a)k$$
$$\Rightarrow (ax+by+cz)+(az+bx+cy)=\{3(ab+bc+ca)-(a^2+b^2+c^2)\}k\tag3$$
$$c(x+y)=c(3a-b)k,\ \ a(y+z)=a(3b-c)k,\ \ b(z+x)=b(3c-a)k$$
$$\Rightarrow (ay+bz+cx)+(az+bx+cy)=\{3(ab+bc+ca)-(ab+bc+ca)\}k\tag4$$
Hence, calculating $(2)+(3)-(4)$ gives us
$$2(ax+by+cz)=2(a^2+b^2+c^2)k\iff \frac{ax+by+cz}{a^2+b^2+c^2}=k\tag5$$
From $(1)$ and $(5)$, we have the conclusion.
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