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 :)