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?

Your solution is wrong because after your first step we need to prove that:
$$6-(a^2+b^2+c^2)\geq3,$$ which is wrong for $a=2$.
We'll prove that $3$ is a minimal value.
Indeed, we need to prove that$$\frac{2(ab+ac+bc)}{abc}-(a^2+b^2+c^2)\geq3$$ or
$$\frac{2(ab+ac+bc)(a+b+c)}{3abc}-\frac{9(a^2+b^2+c^2)}{(a+b+c)^2}\geq3$$ or
$$\sum_{cyc}(2a^4b+2a^4c+6a^3b^2+6a^3c^2-22a^3bc+6a^2b^2c)\geq0$$ or
$$\sum_{cyc}(2a^4b+2a^4c-2a^3b^2-2a^3c^2+8a^3b^2+8a^3c^2-16a^3bc-6a^3bc+6a^2b^2c)\geq0$$ or
$$2\sum_{cyc}(a^4b-a^3b^2-a^2b^3+ab^4)+8\sum_{cyc}c^3(a^2-2ab+b^2)-3abc\sum_{cyc}(a^2-2ab+b^2)\geq0$$ or
$$\sum_{cyc}(a-b)^2(2ab(a+b)+8c^3-3abc)\geq0,$$ which is true by AM-GM:
$$2ab(a+b)+8c^3\geq3\sqrt[3]{(ab(a+b))^2\cdot8c^3}\geq3\sqrt[3]{32}abc>3abc.$$
The equality occurs for $a=b=c=1,$ which says that we got a minimal value.
