Proving an interesting inequality with square roots Let $a,b,c>0$ be real numbers such that $c \geq a \geq b$ and $a^2 \geq bc$.  Show that 
$$\frac{\sqrt{a^2b+b^2c}}{a+c}+\frac{\sqrt{b^2c+c^2a}}{b+a}+\frac{\sqrt{c^2a+a^2b}}{c+b} \geq \frac{\sqrt a +\sqrt b +\sqrt c}{2}.$$
I tried to augment each term on the left using what we have in the hypothesis:
$$\frac{\sqrt{a^2b+b^2c}}{a+c} \geq \frac{\sqrt{a^2b+b^2c}}{2c} \geq \frac{\sqrt{a^2b+b^2a}}{2c}=\frac{\sqrt{ab(a+b)}}{2c}$$
but this kind of inequality doesn't get me anywhere. Could you give me a hint?
 A: Even the following inequality is true for any positives $a$, $b$ and $c$:
$$\sum_{cyc}\frac{\sqrt{a^2b+b^2c}}{a+c}\geq\frac{\sqrt{a}+\sqrt{b}+\sqrt{c}}{\sqrt2}.$$
Indeed, let $a=x^2$, $b=y^2$ and $c=z^2$, where $x$, $y$ and $z$ are positives.
Thus, by Holder
$$\sum_{cyc}\frac{\sqrt{a^2b+b^2c}}{a+c}=\sqrt{\frac{\left(\sum\limits_{cyc}\frac{\sqrt{a^2b+b^2c}}{a+c}\right)^2\sum\limits_{cyc}\frac{b^2(a+c)^2}{a^2+bc}}{\sum\limits_{cyc}\frac{b^2(a+c)^2}{a^2+bc}}}\geq$$
$$\geq \sqrt{\frac{(a+b+c)^3}{\sum\limits_{cyc}\frac{b^2(a+c)^2}{a^2+bc}}}=\sqrt{\frac{(x^2+y^2+z^2)^3}{\sum\limits_{cyc}\frac{y^4(x^2+z^2)^2}{x^4+y^2z^2}}}$$ and it's enough to prove that
$$2(x^2+y^2+z^2)^3\geq(x+y+z)^2\sum\limits_{cyc}\frac{y^4(x^2+z^2)^2}{x^4+y^2z^2},$$
which is obviously true by BW:
https://www.wolframalpha.com/input/?i=2%28x%5E2%2By%5E2%2Bz%5E2%29%5E3-%28x%2By%2Bz%29%5E2%28y%5E4%28z%5E2%2Bx%5E2%29%5E2%2F%28x%5E4%2By%5E2z%5E2%29%2Bz%5E4%28x%5E2%2By%5E2%29%5E2%2F%28y%5E4%2Bx%5E2z%5E2%29%2Bx%5E4%28y%5E2%2Bz%5E2%29%5E2%2F%28z%5E4%2Bx%5E2y%5E2%29%29%2Cx%3Da%2Cy%3Da%2Bu%2Cz%3Da%2Bv
