Find the smallest possible value of an equation, where $a+b+c=3$ We have the positive real numbers $a, b, c$ such that $a+b+c=3$. Find the minimum value of the equation:
$$A=\frac{2-a^3}{a}+\frac{2-b^3}{b}+\frac{2-c^3}{c}$$
I solved it in the following fashion:
$$
\begin{align}
A&=2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-(a^2+b^2+c^2)\\
&\ge 2\cdot\frac{3^2}{a+b+c}-(a^2+b^2+c^2) \quad \textrm{(Andreescu inequality)}\\
&=2\cdot3-(a^2+b^2+c^2)\\
&=6-(a^2+b^2+c^2).
\end{align}$$
However, $9=(a+b+c)^2\ge 3(ab+bc+ac)$, so
$$3\ge ab+bc+ac.$$
From this we have that $A \ge 6-3=3$, where equality is true for $a=b=c=1$.
Due to the total simplicity of my solution, I have difficulties believing that it is correct, despite having checked it thoroughly many times. Could you please tell me if my solution is correct and suggest some alternative solutions?

We have
$$A = 2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-\left[(a+b+c)^2-2(ab+bc+ca)\right]$$
$$=2(ab+bc+ca)\left(\frac{1}{abc}+1\right)-9.$$
Using know inequality $(ab+bc+ca)^2 \geqslant 3abc(a+b+c)$ and the AM-GM inequality, we get
$$A \geqslant 2\sqrt{3abc(a+b+c)}\cdot \frac{2}{\sqrt{abc}}-9=3.$$
Equality occur when $a=b=c=1.$ So $A_{\min} = 3.$
