Find answer of $\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{y+x}=4$

If I had a very shallow question, then I am sorry. $x,y,z\in\mathbb{N}^{+}$ and$$\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{y+x}=4$$find $x,y,z$.

I try with AM-GM, just get$$ \frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{y+x}\geq\frac{3}{2}$$

This means that the equation must have a real solution, but can not be sure there is an integer solution.

Let: $x=ay=abz$, then the equation becomes:$$\frac{ab}{a+b}+\frac{b}{ab+1}+\frac{1}{a^2+ab}=4$$

Which makes the problem become non-homogeneous, and seems to become more difficult. I have no more ideas. Could anyone help me? Thanks a lot.

There are indeed positive integer solutions. The smallest solution with $x, y, z$ in $\mathbb{N}_+$ is

$\small x = 154476802108746166441951315019919837485664325669565431700026634898253202035277999\\ \small y = 36875131794129999827197811565225474825492979968971970996283137471637224634055579\\ \small z = 4373612677928697257861252602371390152816537558161613618621437993378423467772036$

(Source: https://www.quora.com/How-do-you-find-the-positive-integer-solutions-to-frac-x-y+z-+-frac-y-z+x-+-frac-z-x+y-4 )

This solution was found using elliptic curves. Quite cool, eh?

Well, why not? Multiplying through by denominators leads to cubic surface $$ x^3 + y^3 + z^3 - 3 \left(x y^2 + x^2 y + y z^2 + y^2 z + z x^2 + z^2 x \right) - 5xyz = 0 $$ which has integer points $$ (1,1,-1), $$ $$ (11,4,-1), $$ $$ (11,9,-5) $$ and perhaps no others except for permuting and changing all signs. Or multiplying by any common integer factor.

The first triple cannot be used in the original problem, the second and third work fine.