<s>[INST] 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 (*).
 [/INST] 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.
</s>