<s>[INST] Find the minimum value of $A=\frac{2-a^3}{a}+\frac{2-b^3}{b}+\frac{2-c^3}{c}$ Let $a, b$ and $c$ three positive real numbers such that $a+b+c=3$. Find the minimum value of $$A=\frac{2-a^3}{a}+\frac{2-b^3}{b}+\frac{2-c^3}{c}.$$
Here is my attempt. By symmetry we can assume that $a\leq b\leq c$. The function $f(x)=\frac{2-x^3}{x}$ is decreasing and convex on $]0,2^{1/3}]$. So if $c\leq 2^{1/3}$ then $3f(1)=3f(\frac{a+b+c}{3})\leq f(a)+f(b)+f(c)=A$ with equality if $a=b=c=1$. 
If $c>2^{1/3}$ I don't see how to proceed.
 [/INST] Put $S _1=a+b+c=3$, $S_2=ab+bc+ac$ and $S_3=abc$. Then as it was already noticed,  $A=2(\frac{S_2}{S_3}+S_2)-9$. By AM-GM inequality $2(\frac{S_2}{S_3}+S_2)\geq 4\sqrt{\frac{(S_2)^2}{S_3}}$. But Newton's inequality says that $(S_1/3)(S_3/1)=S_3\leq (S_2/3)^2$. It implies $A\geq 3$.
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