<s>[INST] Geometric inequality involving sides of the triangle I was triying to learn geometric inequalities and I got into this problem:
Let $a,b,c$ be the sides of the $\Delta ABC$. Show that:
$$ \left (\frac S R \right)^2 \le \frac 3 8 \left (\frac {ab \sqrt {ab}} {a+b} + \frac {bc \sqrt {bc}} {b+c} + \frac {ca \sqrt {ca}} {c+a} \right) \le \left (\frac {S} {2r} \right)^2,$$ where S is the area of the triangle, R the circumradius and r the inradius.
 [/INST] A left inequality.
Let $a=y+z$, $b=x+z$ and $c=x+y$. Hence, $x>0$, $y>0$ and $z>0$.
By AM-GM $\sum\limits_{cyc}\frac{\sqrt{a^3b^3}}{a+b}\geq\frac{3abc}{\sqrt[3]{\prod\limits_{cyc}(a+b)}}=\frac{9abc}{3\sqrt[3]{\prod\limits_{cyc}(a+b)}}\geq\frac{9abc}{2(a+b+c)}$.
$\frac{S}{R}=\frac{4S^2}{abc}$.
Thus, it remains to prove that $\frac{9abc}{2(a+b+c)}\geq\frac{1}{6}\left(\frac{16S^2}{abc}\right)^2$ or
$27(x+y)^3(x+z)^3(y+z)^3\geq512x^2y^2z^2(x+y+z)^3$.
Since $(x+y)(x+z)(y+z)\geq\frac{8}{9}(x+y+z)(xy+xz+yz)\Leftrightarrow\sum\limits_{cyc}z(x-y)^2\geq0$, 
it remains to prove that $(xy+xz+yz)^3\geq27x^2y^2z^2$, which is true by AM-GM. Done!
For the proof of the right inequality we can use $\frac{\sqrt{ab}}{a+b}\leq\frac{1}{2}$.
Hence, it remains to prove that $(a+b+c)^2\geq3(ab+ac+bc)$, 
which is $\sum\limits_{cyc}(a-b)^2\geq0$. 
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