Can you prove this fractional equality?

[anemone]

Here is this week's POTW:

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

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[anemone]

Congratulations to kaliprasad for his correct solution:

Solution from kaliprasad:

Spoiler

We have

$\dfrac{a+b}{3x-y}= \dfrac{b+c}{3y-z}=\dfrac{c+a}{3z-x}$

Using law of proportionate

if $\dfrac{p}{q} = \dfrac{r}{s} = \dfrac{t}{u}$ then

$\dfrac{p}{q} = \dfrac{r}{s} = \dfrac{t}{u} = \dfrac{p+r+t}{q+s+u}$ we have

$\dfrac{a+b}{3x-y}= \dfrac{b+c}{3y-z}=\dfrac{c+a}{3z-x}= \dfrac{2(a+b+c)}{2(x+y+z)}= \dfrac{a+b+c}{x+y+z}\cdots(1)$ If $\dfrac{p}{q} = \dfrac{r}{s}$

Then $\dfrac{p}{q} = \dfrac{r}{s} = \dfrac{p-r}{q-s}$

We get from

$\dfrac{a+b+c}{x+y+z}=\dfrac{a+b}{3x-y}= \dfrac{a+b+c-(a+b)}{x+y+z-(3x-y)} = \dfrac{c}{z-2x+2y}\cdots(2)$ using law of proportionate

Similarly

$\dfrac{a+b+c}{x+y+z}=\dfrac{b+c}{3y-z}= \dfrac{a+b+c-(b+c)}{x+y+z-(3y-z)} = \dfrac{a}{x-2y+2z}\cdots(3)$

and

$\dfrac{a+b+c}{x+y+z}=\dfrac{c+a}{3z-x}= \dfrac{a+b+c-(c+a)}{x+y+z-(3z-x)} = \dfrac{b}{y-2z+2x}\cdots(4)$

hence from (2) (3) and (4) we get

$\dfrac{a+b+c}{x+y+z}=\dfrac{a}{x-2y+2z}= \dfrac{b}{y-2z+2x} = \dfrac{c}{z-2x+2y}\cdots(5)$Using

$\dfrac{p}{q} = \dfrac{r}{s} = \dfrac{t}{u}= \dfrac{xp+yr+zt}{xq+ys+zu}$

We get

$\dfrac{a}{x-2y+2z}= \dfrac{b}{y-2z+2x} = \dfrac{c}{z-2x+2y}$

$ = \dfrac{ax + by+ cz}{x(x-2y+2z) + y(y-2z+2x) + z(z-2x+2y)}$

= $\dfrac{ax + by+ cz}{x^2+ y^2 + z^2}\cdots(6)$

From (5) and (6) we get

$\dfrac{a+b+c}{x+y+z}=\dfrac{ax+by+cz}{x^2+y^2+z^2}$