If $\sum\limits_{k=1}^n\frac1{x_k+y_k}\leqslant\frac n2$, prove that $\sum\limits_{k=1}^n(\frac{x_k}{1+y_k}+\frac{y_k}{1+x_k})\geqslant n$ 
Let $n \in \mathbb{N}, n\geqslant 2$ and $x_1,x_2,\cdots,x_n,y_1,y_2,\cdots,y_n>0$ with $$\sum_{k=1}^{n}{ 1 \over{x_k+y_k} }  \leqslant {n\over 2}.$$
  Show that $$\sum_{k=1}^{n}\left({ x_k \over{1+y_k} }+{y_k\over{ 1+x_k}}\right)  \geqslant n.$$

I have tried a lot of ways, but I cannot find a good solution. Does anyone have an idea?
 A: For each $k$,\begin{align*}
\frac{x_k}{1 + y_k} + \frac{y_k}{1 + x_k} &= \frac{(x_k + y_k + 1)(x_k + y_k + 2)}{(1 + x_k)(1 + y_k)} - 2\\
&\geqslant \frac{(x_k + y_k + 1)(x_k + y_k + 2)}{\dfrac{1}{4}((1 + x_k) + (1 + y_k))^2} - 2 = \frac{2(x_k + y_k)}{x_k + y_k + 2},
\end{align*}
thus$$
\sum_{k = 1}^n \left( \frac{x_k}{1 + y_k} + \frac{y_k}{1 + x_k} \right) \geqslant \sum_{k = 1}^n \frac{2(x_k + y_k)}{x_k + y_k + 2}.
$$
By Cauchy's inequality,\begin{align*}
&\mathrel{\phantom{=}} \left( \sum_{k = 1}^n \left( 1 + \frac{2}{x_k + y_k} \right) \right) \left( \sum_{k = 1}^n \frac{2(x_k + y_k)}{x_k + y_k + 2} \right)\\
&= \left( \sum_{k = 1}^n \left( 1 + \frac{2}{x_k + y_k} \right) \right) \left( \sum_{k = 1}^n \frac{2}{1 + \frac{2}{x_k + y_k}} \right) \geqslant \left( \sum_{k = 1}^n \sqrt{2} \right)^2 = 2n^2,
\end{align*}
and by condition,$$
\sum_{k = 1}^n \left( 1 + \frac{2}{x_k + y_k} \right) = n + \sum_{k = 1}^n \frac{2}{x_k + y_k} \leqslant 2n,
$$
thus$$
\sum_{k = 1}^n \left( \frac{x_k}{1 + y_k} + \frac{y_k}{1 + x_k} \right) \geqslant \sum_{k = 1}^n \frac{2(x_k + y_k)}{x_k + y_k + 2} \geqslant n.
$$
