Inequality $\sum\frac{x}{(x + n^2)^2}<\frac{1}{2} \sum \frac{1}{x + n^2} $ 
$x\geq0$, then, we have

$$\sum_{n=1}^{\infty}\frac{x}{(x + n^2)^2}<\frac{1}{2} \sum_{n=1}^{\infty} \frac{1}{x + n^2} $$


The problem is not easy, even $x=1$. Any help will be appreciated

The problem is easy for $ 0 \leq x \leq 1$. We have $x \leq n^2$, and hence $ \frac{ x}{x+n^2} \leq \frac{1}{2}$. Thus,
$$ \sum_{n=1}^{\infty} \frac{ x}{(x+n^2)^2} \leq \sum_{n=1}^{\infty} \frac{1}{2} \times \frac{1}{x+n^2}.$$
It remains to verify that we have strict inequality in at least one case.

It is interesting to note that $\int_0^\infty \frac{ x} { (x+y^2)^2} \, dy =\frac{1}{2} \int_0^\infty \frac{1}{x+y^2} \, dy $. This possibly motivates the analysis approach.
