Let $x_{1},\cdots, x_{n}>-1$ be real numbers such that $\sum{x_{i}}=n$. Prove that: $$\sum_{i=1}^{n}{\frac{1}{x_{i}+1}}\geq \sum_{i=1}^{n}{\frac{x_{i}}{x_{i}^{2}+1}}$$
My proof: By AM-HM and $\sum{x_{i}}=n$, LHS $\geq \frac{n^{2}}{n+n\times 1}=\frac{n}{2}$
On the other hand, by using AM-GM in the denominator, RHS $\leq \sum{\frac{x_{i}}{2x_{i}}}=\frac{n}{2} \rightarrow$ LHS $\geq$ RHS, with equality iff for all $i$, $x_{i}=1$.
Is this proof correct?
Your first step is right.
Now, your inequality follows from the following C-S.
$$\sum_{i=1}^n(1+x_i)\sum_{cyc}\frac{1}{1+x_i}\geq n^2.$$
Also, we can use the TL method: $$\sum_{i=1}^n\left(\frac{1}{1+x_i}-\frac{1}{1+x_i^2}\right)=\sum_{i=1}^n\frac{1-x_i}{(1+x_i)(1+x_i^2)}=$$ $$=\sum_{i=1}^n\left(\frac{1-x_i}{(1+x_i)(1+x_i^2)}+\frac{1}{4}(x_i-1)\right)=\sum_{i=1}^n\frac{(x_i-1)^2(x_i^2+2x_i+3)}{4(1+x_i)(1+x_i^2)}\geq0.$$