$x+y+z=xyz$ and $x,y,z>0$. Proof that $\frac{x}{1+x^2}+\frac{2y}{1+y^2}+\frac{3z}{1+z^2}=\frac{xyz(5x+4y+3z)}{(x+y)(y+z)(z+x)}$

Question

$x+y+z=xyz$ and $x,y,z>0$.

Proof that $$\frac{x}{1+x^2}+\frac{2y}{1+y^2}+\frac{3z}{1+z^2}=\frac{xyz(5x+4y+3z)}{(x+y)(y+z)(z+x)}$$

So far, I have managed to reduce $(x+y)(y+z)(z+x)$:

$$(x+y)(y+z)(z+x)=xyz+x^2y+xz^2+x^2z+y^2z+y^2x+yz^2+xyz$$ $$=2xyz+x^2y+xy^2+yz^2+y^2z+x^2z+xz^2$$ $$=2xyz+(x+y)xy+(y+z)yz+(x+z)xz$$ $$=2xyz+(xy-1)xyz+(yz-1)xyz+(xz-1)xyz$$ $$=xyz(xy+yz+zx-1)$$ So the original equation equals to:

$$\frac{x}{1+x^2}+\frac{2y}{1+y^2}+\frac{3z}{1+z^2}=\frac{5x+4y+3z}{xy+yz+zx-1}$$

So I was stuck here: How to simplified the left part of the equation? Also, I have noticed $5x+4y+3z=6(x+y+z)-(x+2y+3z)$ but they seemed to be unhelpful.

Any help is much appreciated, thanks!

Answer 1

Though tedious, this is a purely mechanical problem:

Expanding the denominators, the problem becomes equivalent to the polynomial $$ (x + y) (y + z) (z + x)(x(1 + z^2)(1 + y^2) + 2y(1 + z^2)(1 + x^2) + 3z(1 + y^2)(1 + x^2)) \\ -(1 + z^2)(1 + y^2)(1 + x^2)(x y z (5 x + 4 y + 3 z)) $$ being 0 given $xyz=x+y+z$.

Expand the products and each time you reach a product $xyz$ replace it by $x+y+z$ and keep expanding and cancelling terms. In the end, all terms will cancel.

This is really the job for a computer algebra system of your choice.