Prove: $(a+b) \sqrt {ab}+(a+c) \sqrt {ac}+(b+c)\sqrt {bc}≥ \frac {(a+b+c)^2}{2}$ 
EDİTED:
I tried to solve the last question.I can not get the state of equality and I can not find my mistake. Can you show me my mistake(s)?

4.
  Let, $a,b,c$ be the lengths of sides of a triangle. Prove the inequality:
$$(a+b) \sqrt {ab}+(a+c) \sqrt {ac}+(b+c)\sqrt {bc}≥ \frac {(a+b+c)^2}{2}$$

I will use:
$(1)$ 

$\begin{cases} a+b≥2\sqrt{ab} &\\ b+c≥2\sqrt{bc}&\\ a+c≥2\sqrt{ac} \end{cases}\Longrightarrow \begin{cases} (a+b)\sqrt{ab}≥2ab &\\ (b+c)\sqrt{bc}≥2bc&\\ (a+c)\sqrt{ac}≥2ac \end{cases} \Longrightarrow (a+b)\sqrt{ab}+(b+c)\sqrt{bc}+(a+c)\sqrt{ac}≥2(ab+bc+ac)$

$(2)$

$\begin{cases} b(a+c)>b^2 &\\ c(a+b)>c^2&\\ a(b+c)>a^2 \end{cases}\Longrightarrow 2(ab+bc+ac)>a^2+b^2+c^2 \Longrightarrow \frac{a^2+b^2+c^2}{2}<ab+bc+ac$

Applying $(1)$ and $(2)$ we have

$(a+b) \sqrt {ab}+(a+c) \sqrt {ac}+(b+c)\sqrt {bc}- \frac {(a+b+c)^2}{2}≥2ab+2bc+2ac- \frac {(a+b+c)^2}{2}≥ab+bc+ac-\frac{a^2+b^2+c^2}{2}>0≠≥0$


Let $a=y+z$, $b=x+z$ and $c=x+y$.
Hence, $x$, $y$ and $z$ are positives and we need to prove that
$$\sum_{cyc}(2x+y+z)\sqrt{(x+y)(x+z)}\geq2(x+y+z)^2$$ and since by C-S $$\sqrt{(x+y)(x+z)}\geq x+\sqrt{yz},$$ it's enough to prove that
$$\sum_{cyc}(2x+y+z)(x+\sqrt{yz})\geq2(x+y+z)^2$$ or
$$\sum_{cyc}\left(\sqrt{x^3y}+\sqrt{x^3z}-2xy+2x\sqrt{yz}\right)\geq0$$ or
$$\sum_{cyc}\left(\sqrt{xy}(\sqrt{x}-\sqrt{y})^2+2x\sqrt{yz}\right)\geq0.$$
Done!
The equality does not occur because $\sum\limits_{cyc}x\sqrt{yz}>0.$ 
