Inequality. $\sum{(a+b)(b+c)\sqrt{a-b+c}} \geq 4(a+b+c)\sqrt{(-a+b+c)(a-b+c)(a+b-c)}.$ Let $a,b,c$ be the side-lengths of a triangle. Prove that: 
I.
$$\sum_{cyc}{(a+b)(b+c)\sqrt{a-b+c}} \geq 4(a+b+c)\sqrt{(-a+b+c)(a-b+c)(a+b-c)}.$$
What I have tried: 
\begin{eqnarray}
a-b+c&=&x\\
b-c+a&=&y\\
c-a+b&=&z.
\end{eqnarray}
So $a+b+c=x+y+z$ and $2a=x+y$, $2b=y+z$, $2c=x+z$ and our inequality become:
$$\sum_{cyc}{\frac{\sqrt{x}\cdot(x+2y+z)\cdot(x+y+2)}{4}} \geq 4\cdot(x+y+z)\cdot\sqrt{xyz}. $$ Or if we make one more notation : $S=x+y+z$ we obtain that:
$$\sum_{cyc}{\sqrt{x}(S+y)(S+z)} \geq 16S\cdot \sqrt{xyz} \Leftrightarrow$$
$$S^2(\sqrt{x}+\sqrt{y}+\sqrt{z})+S(y\sqrt{x}+z\sqrt{x}+x\sqrt{y}+z\sqrt{y}+x\sqrt{z}+y\sqrt{z})+xy\sqrt{z}+yz\sqrt{x}+xz\sqrt{y} \geq 16S\sqrt{xyz}.$$ To complete the proof we have to prove that: 
$$y\sqrt{x}+z\sqrt{x}+x\sqrt{y}+z\sqrt{y}+x\sqrt{z}+y\sqrt{z} \geq 16\sqrt{xyz}. $$
Is this last inequality true ? 
II.
Knowing that: $$p=\frac{a+b+c}{2}$$ we can rewrite the inequality: 
$$\sum_{cyc}{(2p-c)(2p-a)\sqrt{2(p-b)}} \geq 8p \sqrt{2^3 \cdot (p-a)(p-b)(p-c)} \Leftrightarrow$$
$$\sum_{cyc}{(2p-c)(2p-a)\sqrt{(p-b)}} \geq 16p \sqrt{(p-a)(p-b)(p-c)}$$
This help me ?
Thanks :) 

Notice that the inequality proposed is proved once we estabilish

$$\frac{(a+b)(b+c)}{\sqrt{a+b-c}\sqrt{-a+b+c}}+\frac{(b+c)(c+a)}{\sqrt{a-b+c}\sqrt{-a+b+c}}+\frac{(c+a)(a+b)}{\sqrt{a-b+c}\sqrt{a+b-c}}\geq 4(a+b+c).$$

Using AM-GM on the denominators in the LHS, we estabilish that 


*

*$\sqrt{a+b-c}\sqrt{-a+b+c}\leq b$,

*$\sqrt{a-b+c}\sqrt{-a+b+c}\leq c$,

*$\sqrt{a-b+c}\sqrt{a+b-c}\leq a$.


Therefore 

$$\operatorname{LHS}\geq \frac{(a+b)(b+c)}{b}+\frac{(b+c)(c+a)}{c}+\frac{(c+a)(a+b)}{a}=3(a+b+c)+\frac{ca}{b}+\frac{ab}{c}+\frac{bc}{a}$$

To finish with the proof it suffices then to prove 

$$\frac{ca}{b}+\frac{ab}{c}+\frac{bc}{a}\geq a+b+c.$$

This follows from AM-GM, indeed


*

*$$\frac{\frac{ac}{b}+\frac{ab}{c}}{2}\geq a,$$

*$$\frac{\frac{bc}{a}+\frac{ab}{c}}{2}\geq b,$$

*$$\frac{\frac{ac}{b}+\frac{bc}{a}}{2}\geq c;$$


summing up these last inequalities, we get the desired result, hence finishing the proof. Notice that the condition of $a,b,c$ being the sides of a triangle is essential in the first usage of AM-GM.
