Prove that $\sum \frac{x}{x^2+7}\le \frac{3}{8}$

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 $f\geq 0$.

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 $t \in (0, \frac\pi2)$.