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 $$x_1+x_2+\cdot\cdot\cdot+x_n= \sqrt{n}$$ $$x_1,x_2,\cdot\cdot\cdot,x_n \ge0$$ $$ n \in \mathbb{N}$$

I'm afraid the inequality is wrong. Note that RHS $< 1$. However, if we take $x:=1$, then $x^2/(x^2+1)^2=1/4$. Thus, we choose a large $n$, and let as much $x_i$ as possible be $1$, then the inequality fails. But I believe that there exists a bound of $n$ to let the inequality hold.
