How to prove $\sqrt{a+b}\sqrt{b+c}+\sqrt{b+c}\sqrt{c+a}+\sqrt{c+a}\sqrt{a+b}\geq \sqrt{3(ab+bc+ca)}+(a+b+c)$? Recently I  meet a problem ,it says

Suppose $a,b,c,x,y,z\in \mathbb{R}^+$,then 
  \begin{align*}
\frac{x}{y+z}(b+c)+\frac{y}{z+x}(a+c)+\frac{z}{x+y}(a+b)\geq
\sqrt{3(ab+bc+ca)} 
\end{align*}

Fix $a,b,c$,then the original inequality is equal to 
\begin{align*}
\frac{x+y+z}{y+z}(b+c)+\frac{x+y+z}{z+x}(a+c)+\frac{x+y+z}{x+y}(a+b)\geq \sqrt{3(ab+bc+ca)}+2(a+b+c)
\end{align*}
By using Cauchy's inequality,we  can get 
\begin{align*}
\frac{x+y+z}{y+z}(b+c)+\frac{x+y+z}{z+x}(a+c)+\frac{x+y+z}{x+y}(a+b)\geq \frac{1}{2}(\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a})^2
\end{align*}
So if we can proof (Since the original equality is true ,then the following equality must be true)
\begin{align*}
(\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a})^2\geq 2\sqrt{3(ab+bc+ca)}+4(a+b+c)
\end{align*}
or
\begin{align*}
\sqrt{a+b}\sqrt{b+c}+\sqrt{b+c}\sqrt{c+a}+\sqrt{c+a}\sqrt{a+b}\geq \sqrt{3(ab+bc+ca)}+(a+b+c)\tag{*}
\end{align*}
then the problem is done.But I can't prove (*).
 A: Solution of AOPS member (don't remember name)
Setting
$$x=\sqrt{(a+b)(a+c)}-a, \quad y=\sqrt{(b+c)(b+a)}-b, \quad z=\sqrt{(c+a)(c+b)}-c,$$
we have
$$ab+bc+ca=xy+yz+zx,$$
Inequality become
$$x+y+z \geqslant \sqrt{3(xy+yz+zx)}.$$
Done.
A: We'll prove a more general result:
If $x,y,z$ are the sides of a triangle, then
$$2\left(xy+yz+zx\right)\geq4S\sqrt3+x^2+y^2+z^2.$$
Indeed, as the $RHS$ is a decreasing function of $xyz,$ then according to the $uvw$ principles for triangle sides, we only need to check in the latter $y=z=1$ and $x\in[0,1]$ and get
$$x^2\left(x-1\right)^2\geq0,$$
which is obvious. We are done.
P.S. I just realised that I proved Finsler–Hadwiger theorem.
A: Let $ab+ac+bc=1$.
Thus, we need to prove that:
$$\sum_{cyc}\sqrt{a^2+1}\geq\sqrt3+a+b+c.$$
Now, let $a=\tan x$, $b=\tan y$ and $c=\tan z$, where $\{x,y,z\}\subset\left(0^{\circ},90^{\circ}\right).$
Thus, $x+y+z=90^{\circ}$ and we need to prove that $$\sum_{cyc}f(x)\geq\sqrt3,$$ where $$f(x)=\frac{1}{\cos{x}}-\tan{x}.$$ 
But $$f''(x)=\frac{(1-\sin{x})^2}{\cos^3x}>0$$ and now our inequality follows from Jensen.
