Prove that $\sum \frac{x}{x^2+7}\le \frac{3}{8}$ Let $x,y,z>0$ such that $xy+yz+xz=3$. Show that $$\frac{x}{x^2+7}+\frac{y}{y^2+7}+\frac{z}{z^2+7}\le \frac{3}{8}$$

We have: $$x+y+z\ge \sqrt{3\left(xy+yz+xz\right)}=3\rightarrow \frac{3}{8\left(x+y+z\right)}\le \frac{3}{8\cdot 3}$$
Then i will prove  $$\sum \frac{x}{3x^2+7\left(xy+xz+yz\right)}\le \frac{3}{8\left(x+y+z\right)}$$
But it's wrong. I tried to use uvw and done. So i need another idead without uvw

Hint: Use the substitution $x = \sqrt 3 \tan A$ and so on for some acute $\triangle ABC$, and then Jensen’s inequality as $t \mapsto \dfrac{\sqrt 3 \tan t}{3\tan^2t + 7}$ is concave for $t \in (0, \frac\pi2)$.
