Prove that if $\frac{x+y}{3a-b}=\frac{y+z}{3b-c}=\frac{z+x}{3c-a}$ then $\frac{x+y+z}{a+b+c}=\frac{ax+by+cz}{a^2+b^2+c^2}$

Question

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?

Answer 1

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.