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?
 A: 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.
A: 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 $A_{\min} = 3.$
