<s>[INST] How prove $\sum\limits_{cyc} \frac{ab}{c} \geq 3.$ Let $a,b,c>0 ,\sum\limits_{cyc} a^{3}=3$. How prove $\sum \frac{ab}{c} \geq 3.$ 
My try:
$ a^3+b^3+c^3=3$
$a^2b^2+a^2c^2+b^2c^2 \ge 3abc$
so
$ (a+b+c)^3-3(a+b+c)(ab+ac+bc)+3abc=3$
$(ab+ac+bc)^2-2(a+b+c)abc \ge 3abc$
 [/INST] The inequality is false: take $a=\frac{1}{\sqrt[3] 6}$,$b=\frac{1}{\sqrt[3] 6}$,$c=\frac{\sqrt[3] 8}{\sqrt[3] 3}$, you obtain
$$ \sim 2.9918... \ge 3.$$
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