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
 A: 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.
A: From a purely algebraic point of view, this problem is quite interesting if we first notice that $$\frac {d}{dx} \Big( \frac {1}{x+n^2} \Big)=-\frac{1}{\left(x+n^2\right)^2}$$ The second point is to recognize that
$$ \sum_{n=1}^{\infty} \frac{1}{x + n^2}=\frac{\pi  \sqrt{x} \coth \left(\pi  \sqrt{x}\right)-1}{2 x}$$ So $$\sum_{n=1}^{\infty}\frac{x}{(x + n^2)^2}-\frac{1}{2} \sum_{n=1}^{\infty} \frac{1}{x + n^2}=\frac{\pi ^2 x \text{csch}^2\left(\pi  \sqrt{x}\right)-1}{4 x}$$ which is always negative.
A: I think this approach is also valid, and simpler.
A term-wise subtraction (LHS - RHS) gives:
$$
\frac{x}{(x+n^2)^2} - \frac{1}{2(x+n^2)} = \frac{2x - (x+n^2)}{2(x+n^2)^2} = \frac{x-n^2}{2(x+n^2)^2}
$$
for sufficiently large number of terms, this can be proved to be less than zero.
