<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] $\dfrac{2-a^3}{a}+\dfrac{2-b^3}{b}=\dfrac{2(a+b)}{ab}+2ab-(a+b)^2$
let $c$ is max $, a+b \le 2 ,x=ab \le \dfrac{(a+b)^2}{4} \le 1$, consider $f(x)=\dfrac{a+b}{x}+x, a+b \ge 2\sqrt{ab}\ge 2ab >ab=x\ge x^2 \implies f(x)$ will get min when $x$ get max.
so $A$ will get min when $ab=\dfrac{(a+b)^2}{4}=\dfrac{(3-c)^2}{4}$  
$A \ge \dfrac{8}{3-c}-\dfrac{(3-c)^2}{2}+\dfrac{2}{c}-c^2=g(c) $
$g'(c)=(1-c)(1-\dfrac{2(c+3)}{(3-c)^2c^2}), (3-c)c \le \dfrac{(3-c+c)^2}{4}=\dfrac{9}{4}< \sqrt{2(c+3)}, g'(c)=0 $ will have only solution $c=1 $ ,it is easy to verify $g(1)=3$ is min
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