<s>[INST] Nice olympiad inequality Let's go for an olympiad inequality :

let $a,b,c>0$ then we have :
  $$\sum_{cyc}\frac{ab}{a+b}\geq \frac{3\sqrt{3}}{2}\sqrt{\frac{abc}{a+b+c}}$$

My proof :
$$\sum_{cyc}\frac{ab}{a+b}=\frac{(a^2 b^2 + 3 a^2 b c + a^2 c^2 + 3 a b^2 c + 3 a b c^2 + b^2 c^2)}{(a+b)(b+c)(c+a)}=\frac{(ab  + ac  + bc)^2+abc(a+b+c)}{(ab+bc+ca)(a+b+c)-abc}$$
So if we use the $uwv's$ method we get :
$$\frac{9v^4+w^3(3u)}{3v^2(3u)-w^3}\geq \frac{3\sqrt{3}}{2}\sqrt{\frac{w^3}{3u}}$$
But I can't conclue by myself so can you help me ? 
 [/INST] By your work we need to prove that $f(w^3)\geq0,$ where
$$f(w^3)=3v^4+uw^3+\frac{1}{2}(w^3-9uv^2)\sqrt{\frac{w^3}{u}}.$$
We see that $f$ increases, which says that it's enough to prove our inequality for a minimal value of $w^3$, which happens in the following cases.


*

*$w^3\rightarrow0^+$.


In this case our inequality is obvious;


*Two variable are equal.


Since our inequality is homogeneous and symmetric, it's enough to assume $b=c=1$,
which gives
$$\frac{5a+1}{a+1}\geq3\sqrt3\sqrt{\frac{a}{a+2}},$$ which is wrong for $a\rightarrow+\infty.$
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