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
 A: Also, the Tangent Line method helps.
Let $x=\sqrt3\tan\alpha,$ $y=\sqrt3\tan\beta$ and $z=\sqrt3\tan\gamma,$ where $\{\alpha,\beta,\gamma\}\subset\left(0,\frac{\pi}{2}\right).$
Thus, $$\sum_{cyc}\tan\alpha\tan\beta=1.$$ It follows that $$\gamma=\arctan\left(\frac{1-\tan(\alpha)\tan(\beta)}{\tan\alpha+\tan\beta}\right)=\arctan(\cot(\alpha+ \beta))=\arctan\left(\tan\left(\frac\pi2-\alpha-\beta\right)\right),$$ which gives
$$\alpha+\beta+\gamma=\frac{\pi}{2}$$ and we obtain:
\begin{split}\frac{3}{8}-\sum_{cyc}\frac{x}{x^2+7}&=\sqrt3\sum_{cyc}\left(\frac{1}{8\sqrt3}-\frac{\tan\alpha}{3\tan^2\alpha+7}\right)\\
&=\sqrt3\sum_{cyc}\left(\frac{1}{8\sqrt3}-\frac{\tan\alpha}{3\tan^2\alpha+7}+\frac{1}{8}\left(\alpha-\frac{\pi}{6}\right)\right)\geq0.\end{split}
Proof of the last inequality: For $x\in\left[0,\frac{\pi}{2}\right]$, let
$$f(x)= \frac{1}{8\sqrt3}-\frac{\tan x}{3\tan^2 x+7}+\frac{1}{8}\left(x-\frac{\pi}{6}\right).$$
Then $$f'(x)=\frac18+\frac{\sec^2(x)\cdot (-7 + 3 \tan^2(x))}{(7 + 3 \tan^2(x))^2}=\frac18-\frac{(5\cos(2x)+2)\cdot\sec^4(x)}{\big((2\cos(2x)+5)\cdot\sec^2(x)\big)^2}=\frac18-\frac{5\cos(2x)+2}{(2\cos(2x)+5)^2}.$$
Hence, for $x\in(0,\pi/2)$, we have $f'(x)=0$ iff $x=\frac\pi6$. Since $f(\pi/6)=0$ and $f(0)>0$, $f(\pi/2)>0$, we conclude $f\geq 0$.
A: 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 $t \in (0, \frac\pi2)$.
