Did you mean $yz$ in the numerator of the second fraction? Why would it be so almost symmetric?

if I understood correctly the given solution, because ineq is not my thing, it has typos, last numerator should be $y^2$: Quote: Let $x, y, z$ be positive reals. Prove that $$\frac{xz}{x^2 + xy + y^2 + 6z^2} + \frac{zx}{z^2 + zy + y^2 + 6x^2} + \frac{y^2}{x^2 + xz + z^2 + 6y^2} \le \frac13$$