<s>[INST] Proving $\sum_{cyc}\frac{\sqrt{s(s-a)}}{a}\geq\frac{3\sqrt{3}}2$ for $a$, $b$, $c$, $s$ the sides and semi-perimeter of a triangle 
If $a, b, c$ are the lengths of the sides of a triangle and s is the semiperimeter, prove that:
$$\sum_{cyc}\frac{\sqrt{s(s-a)}}{a}\geq \frac{3\sqrt{3}}{2}$$

My attempt: $$\sum_{cyc} cos \,\frac{A}{2}=\sum _{cyc} \sqrt{\frac{s(s-a)}{bc}}\leq \frac{3 \sqrt{3}}{2}\leq \sum_{cyc}\frac{\sqrt{s(s-a)}}{a}$$. As the first inequality is obvious, it is enough to show that $$\sum _{cyc} \sqrt{\frac{s(s-a)}{bc}}\leq \sum_{cyc}\frac{\sqrt{s(s-a)}}{a}$$. We prove that $$\sqrt{\frac{s(s-a)}{bc}} \leq\frac{\sqrt{s(s-a)}}{a}$$ or equivalently $a^{2}\leq bc$. Similary $b^{2}\leq ac$ and $c^{2}\leq ab$ and adding the inequalities we get $(a-b)^{2}+(b-c)^{2}+ (c-a)^{2}\leq 0$.
 [/INST] Let $a = y+z,\,b = z+x,\,c = x+y$ for $x,\,y,\,z>0,$ we have
$$\sum \frac{\sqrt{s(s-a)}}{a} = \sum \frac{\sqrt{\frac{a+b+c}{2}\left(\frac{a+b+c}{2}-a\right)}}{a} = \sum \frac{\sqrt{x(x+y+z)}}{y+z}.$$
We will show that
$$\sum \frac{\sqrt{x(x+y+z)}}{y+z} \geqslant \frac{3\sqrt 3}{2}.$$
Indeed, suppose $x+y+z=3$ then the inequality become
$$\sum \frac{\sqrt{x}}{3 - x} \geqslant \frac{3}{2}.$$
We have
$$\left(\frac{\sqrt{x}}{3 - x}\right)^2 - \left(\frac{x}{2}\right)^2  = \frac{x(4 - x)(x-1)^2}{4(3-x)^2} \geqslant 0.$$
So
$$ \frac{\sqrt{x}}{3 - x} \geqslant \frac{x}{2},$$
or
$$ \sum \frac{\sqrt{x}}{3 - x} \geqslant \frac{x+y+z}{2} = \frac32.$$
The proof is completed.
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