Let $a, b, c$ be positive real numbers such that $a+b+c=3$. Find the minimum value of the expression

$$
A=\frac{2-a^{3}}{a}+\frac{2-b^{3}}{b}+\frac{2-c^{3}}{c}
$$

![](https://cdn.mathpix.com/cropped/2024_06_05_f56efd4e6fb711c0f78eg-2.jpg?height=348&width=477&top_left_y=109&top_left_x=241)

$19^{\text {th }}$ Junior Balkan Mathematical Olympiad June 24-29, 2015, Belgrade, Serbia
We can rewrite $A$ as follows:

$$
\begin{aligned}
& A=\frac{2-a^{3}}{a}+\frac{2-b^{3}}{b}+\frac{2-c^{3}}{c}=2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-a^{2}-b^{2}-c^{2}= \\
& 2\left(\frac{a b+b c+c a}{a b c}\right)-\left(a^{2}+b^{2}+c^{2}\right)=2\left(\frac{a b+b c+c a}{a b c}\right)-\left((a+b+c)^{2}-2(a b+b c+c a)\right)= \\
& 2\left(\frac{a b+b c+c a}{a b c}\right)-(9-2(a b+b c+c a))=2\left(\frac{a b+b c+c a}{a b c}\right)+2(a b+b c+c a)-9= \\
& 2(a b+b c+c a)\left(\frac{1}{a b c}+1\right)-9
\end{aligned}
$$

Recall now the well-known inequality $(x+y+z)^{2} \geq 3(x y+y z+z x)$ and set $x=a b, y=b c, z=c a$, to obtain $(a b+b c+c a)^{2} \geq 3 a b c(a+b+c)=9 a b c$, where we have used $a+b+c=3$. By taking the square roots on both sides of the last one we obtain:

$$
a b+b c+c a \geq 3 \sqrt{a b c}
$$

Also by using AM-GM inequality we get that

$$
\frac{1}{a b c}+1 \geq 2 \sqrt{\frac{1}{a b c}}
$$

Multiplication of (1) and (2) gives:

$$
(a b+b c+c a)\left(\frac{1}{a b c}+1\right) \geq 3 \sqrt{a b c} \cdot 2 \sqrt{\frac{1}{a b c}}=6
$$

So $A \geq 2 \cdot 6-9=3$ and the equality holds if and only if $a=b=c=1$, so the minimum value is 3.