Prove that $\sum\limits_{cyc}\frac{a-b}{\sqrt{b+c}}\geq0$ 
Let $a$, $b$, $c$ and $d$ be non-negative numbers such that $\prod\limits_{cyc}(a+b)\neq0$. Prove that:
  $$\frac{a-b}{\sqrt{b+c}}+\frac{b-c}{\sqrt{c+d}}+\frac{c-d}{\sqrt{d+a}}+\frac{d-a}{\sqrt{a+b}}\geq0$$

The equality occurs also for $a=c$ and $b=d$.
My trying.
We need to prove that
$$\sum_{cyc}\frac{a+c-b-c}{\sqrt{b+c}}\geq0$$ or
$$(a+c)\left(\frac{1}{\sqrt{b+c}}+\frac{1}{\sqrt{d+a}}\right)+(b+d)\left(\frac{1}{\sqrt{c+d}}+\frac{1}{\sqrt{a+b}}\right)\geq\sum_{cyc}\sqrt{a+b}$$
and what is the rest?
Thank you!
 A: Since the sign of $\frac{a-b}{\sqrt{b+c}}+\frac{b-c}{\sqrt{c+d}}+\frac{c-d}{\sqrt{d+a}}+\frac{d-a}{\sqrt{a+b}}$ is preserved when $(a,b,c,d)$ is replaced with $(ax,bx,cx,dx)$ for any positive $x$, we can assume that $a+b+c+d=1$ WLOG.
Now let $x=a+b,y=a+c,z=a+d$ and from OP's last inequality, we get $$y\left(\frac{1}{\sqrt {z}}+\frac{1}{\sqrt {1-z}}\right)+(1-y)\left(\frac{1}{\sqrt {x}}+\frac{1}{\sqrt {1-x}}\right) \ge \sqrt x+\sqrt {1-x}+\sqrt z+\sqrt {1-z}$$
To prove this, let's consider this inequality.$$\frac{1}{\sqrt {x}}+\frac{1}{\sqrt {1-x}} \ge \sqrt x+\sqrt {1-x}+\sqrt z+\sqrt {1-z}$$
And $\sqrt z+\sqrt {1-z}$ is maximized at $\sqrt 2$ when $z=1/2$.
$$\frac{1}{\sqrt {x}}+\frac{1}{\sqrt {1-x}} \ge \sqrt x+\sqrt {1-x}+\sqrt 2$$
We can verify this by using calculus, graphing tools, etc.
Therefore, $\frac{1}{\sqrt {x}}+\frac{1}{\sqrt {1-x}} \ge \sqrt x+\sqrt {1-x}+\sqrt z+\sqrt {1-z}$ and similarly $$\frac{1}{\sqrt {z}}+\frac{1}{\sqrt {1-z}} \ge \sqrt x+\sqrt {1-x}+\sqrt z+\sqrt {1-z}$$
Now we get $$\begin{align}&y\left(\frac{1}{\sqrt {z}}+\frac{1}{\sqrt {1-z}}\right)+(1-y)\left(\frac{1}{\sqrt {x}}+\frac{1}{\sqrt {1-x}}\right) \\\ge& y(\sqrt x+\sqrt {1-x}+\sqrt z+\sqrt {1-z})+(1-y)(\sqrt x+\sqrt {1-x}+\sqrt z+\sqrt {1-z})\\=&\sqrt x+\sqrt {1-x}+\sqrt z+\sqrt {1-z}\end{align}$$
Done!
