<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] After using Ravi's substituition $a=x+y,b=y+z,c=x+z$ as nguyenhuyen_ag did we have to prove $$\sum \frac{\sqrt{z(x+y+z)}}{y+x} \geq \frac{3\sqrt 3}{2}.$$  Now we will use the method of Isolated fudging. We guess that we might be able to prove the follwing inequality $$\frac{\sqrt{z(x+y+z)}}{y+x}\ge \frac{3\sqrt{3}}{2}\frac{z}{x+y+z}$$ Indeed after squaring this inequality is equivalent to   $$\frac{{(x+y+z)}^3}{27}\ge z(\frac{x+y}{2})(\frac{x+y}{2})$$ which is evident by AM_GM. Hence $$\sum \frac{\sqrt{x(x+y+z)}}{y+z} \ge\frac{3\sqrt{3}}{2} \sum \frac{z}{x+y+z}=\frac{3\sqrt{3}}{2}$$ done
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