If $\sum x_k= \sqrt{n}$, then $\sum\frac{x_k^2}{(x_k^2+1)^2}\leq\frac{n^2}{(n+1)^2}$ A new question has emerged after this one was successfully answered by r9m: If $a+b+c+d = 2$, then $\frac{a^2}{(a^2+1)^2}+\frac{b^2}{(b^2+1)^2}+\frac{c^2}{(c^2+1)^2}+\frac{d^2}{(d^2+1)^2}\le \frac{16}{25}$. I thought of this generalization. Does it hold?
$$\dfrac{x_1^2}{(x_1^2+1)^2}+\dfrac{x_2^2}{(x_2^2+1)^2}+\cdot\cdot\cdot+\dfrac{x_n^2}{(x_n^2+1)^2}\le \dfrac{n^2}{(n+1)^2}$$ with $$x_1+x_2+\cdot\cdot\cdot+x_n= \sqrt{n}$$ $$x_1,x_2,\cdot\cdot\cdot,x_n \ge0$$ $$ n \in \mathbb{N}$$

Alternative way to reach the result, that is already proposed in the other answers:
Consider the non-linear maximization problem $$\max_{x_i}\sum_{i=1}^{n}\frac{x_i^2}{(x_i^2+1)^2}$$ subject to $x_1+x_2+\ldots+x_n=\sqrt{n}$ and $x_i\ge0$. In that case the Lagrange function is defined as $$Λ(x, λ)=\sum_{i=1}^{n}\frac{x_i^2}{(x_i^2+1)^2}+λ\left(\sum_{i=1}^{n}x_i-\sqrt{n}\right)$$ The equation $\nablaΛ=0$ yields $$λ=\frac{2x_i(1-x_i^2)}{(1+x_i^2)^3}$$ for all $1\le x_i \le n$. Obviously the point $x^0$ with $$x^0_i=\begin{cases}1, & 1\le i \le \lfloor \sqrt{n}\rfloor \\ \\0, & \lfloor \sqrt{n}\rfloor \le i \le n  \end{cases}, \qquad λ=0$$ solves the above system (both sides are equal to zero) and is thus a stationary point. The value of the objective function in $x^0$ is equal to $$\lfloor \sqrt{n}\rfloor\frac14$$
It is easy to see that for $n \ge 16$ this values is greater than $1$ and thus larger than the RHS of the given inequality. For $n<16$ this point does not yield a value larger than the RHS.

Perhaps there are other stationary points that yield a higher value for the objective function, even for $n<16$.
