Prove Alexander Kovaćec inequality 
Let $x_{1},x_{2},\cdots,x_{n}>0$,show that
  $$n\left(\dfrac{3-2\sqrt{2}}{2\sqrt{2}}\right)\min_{1\le i\le n}\sqrt{x_{i}}
+\left(1-\dfrac{1}{2\sqrt{2}}\right)\sum_{i=1}^{n}\sqrt{x_{i}}\le\sum_{i=1}^{n}\dfrac{x_{i}}{\sqrt{x_{i}+x_{i+1}}},x_{n+1}=x_{1}$$

It is said this inequality is proved by A.Kovacec. So I was looking for it ,and maybe the proof is in this paper:Two Contributions to Inequalities by Alexander Kovaćec. But unfortunately I dont't have permission to open this article, so I don't know if this question comes from this article at all. What I am more concerned about is how to prove the inequality.
 A: I believe the following would help.
After replacing $x_{i}\rightarrow x_{i}^2$ we need to prove that
$$\sum_{i=1}^n\frac{x_i^2}{\sqrt{x_i^2+x_{i+1}^2}}\geq\frac{(2\sqrt2-1)\sum\limits_{i=1}^nx_i}{2\sqrt2}+\frac{(3-2\sqrt2)n\min\limits_i{x_i}}{2\sqrt2}$$ or
$$\sum_{i=1}^n\left(\frac{x_i^2}{\sqrt{x_i^2+x_{i+1}^2}}-\frac{x_i}{\sqrt2}\right)\geq\frac{(2\sqrt2-3)\sum\limits_{i=1}^nx_i}{2\sqrt2}+\frac{(3-2\sqrt2)n\min\limits_i{x_i}}{2\sqrt2}$$ or
$$\sum_{i=1}^n\frac{x_i\left(\sqrt2x_i-\sqrt{x_i^2+x_{i+1}^2}\right)}{\sqrt{x_i^2+x_{i+1}^2}}+\frac{3-2\sqrt2}{2}\left(\sum\limits_{i=1}^nx_i-n\min\limits_i{x_i}\right)\geq0$$ or
$$\sum_{i=1}^n\frac{x_i(x_i^2-x_{i+1}^2)}{\sqrt{x_i^2+x_{i+1}^2}\left(\sqrt2x_i+\sqrt{x_i^2+x_{i+1}^2}\right)}+\frac{3-2\sqrt2}{2}\left(\sum\limits_{i=1}^nx_i-n\min\limits_i{x_i}\right)\geq0$$ or
$$\sum_{i=1}^n\left(\frac{x_i(x_i^2-x_{i+1}^2)}{\sqrt{x_i^2+x_{i+1}^2}\left(\sqrt2x_i+\sqrt{x_i^2+x_{i+1}^2}\right)}-\frac{x_i-x_{i+1}}{2}\right)\geq\frac{2\sqrt2-3}{2}\left(\sum\limits_{i=1}^nx_i-n\min\limits_i{x_i}\right)$$ or
$$\sum_{i=1}^n\frac{(x_i-x_{i+1})\left(x_i^2-x_{i+1}^2+2x_ix_{i+1}-x_i\sqrt{2(x_i^2+x_{i+1}^2)}\right)}{\sqrt{x_i^2+x_{i+1}^2}\left(\sqrt2x_i+\sqrt{x_i^2+x_{i+1}^2}\right)}+$$
$$+(3-2\sqrt2)\left(\sum\limits_{i=1}^nx_i-n\min\limits_i{x_i}\right)\geq0$$ or
$$\sum_{i=1}^n\frac{(x_i-x_{i+1})^2\left(x_i+x_{i+1}-\frac{\sqrt2x_i(x_i+x_{i+1})}{\sqrt2x_{i+1}+\sqrt{x_i^2+x_{i+1}^2}}\right)}{\sqrt{x_i^2+x_{i+1}^2}\left(\sqrt2x_i+\sqrt{x_i^2+x_{i+1}^2}\right)}+(3-2\sqrt2)\left(\sum\limits_{i=1}^nx_i-n\min\limits_i{x_i}\right)\geq0$$ or
$$\sum_{i=1}^n\tfrac{(x_i-x_{i+1})^2(x_i+x_{i+1})\left(1-\frac{\sqrt2x_i}{\sqrt2x_{i+1}+\sqrt{x_i^2+x_{i+1}^2}}\right)}{\sqrt{x_i^2+x_{i+1}^2}\left(\sqrt2x_i+\sqrt{x_i^2+x_{i+1}^2}\right)}+(3-2\sqrt2)\left(\sum\limits_{i=1}^nx_i-n\min\limits_i{x_i}\right)\geq0.$$
Maybe from here you'll end it.
By the way, I have a proof for the following.
$$\sum_{i=1}^n\tfrac{(x_i-x_{i+1})^2(x_i+x_{i+1})\left(1-\frac{\sqrt2x_i}{\sqrt2x_{i+1}+\sqrt{x_i^2+x_{i+1}^2}}\right)}{\sqrt{x_i^2+x_{i+1}^2}\left(\sqrt2x_i+\sqrt{x_i^2+x_{i+1}^2}\right)}+0.42\left(\sum\limits_{i=1}^nx_i-n\min\limits_i{x_i}\right)\geq0.$$
For $n=3$ the following stronger inequality is true.
Let $a$, $b$ and $c$ be positive numbers. Prove that:
$$\frac{a}{\sqrt{a+b}}+\frac{b}{\sqrt{b+c}}+\frac{c}{\sqrt{c+a}}\geq\frac{\sqrt{a}+\sqrt{b}+\sqrt{c}}{\sqrt2}$$
