Question 28
Let $\alpha, \beta, \gamma$ be the angles of a triangle opposite to its sides with lengths $a, b$ and $c$, respectively. Prove the inequality
$$
a \cdot\left(\frac{1}{\beta}+\frac{1}{\gamma}\right)+b \cdot\left(\frac{1}{\gamma}+\frac{1}{\alpha}\right)+c \cdot\left(\frac{1}{\alpha}+\frac{1}{\beta}\right) \geq 2 \cdot\left(\frac{a}{\alpha}+\frac{b}{\beta}+\frac{c}{\gamma}\right)
$$
$$
a \cdot\left(\frac{1}{\beta}+\frac{1}{\gamma}\right)+b \cdot\left(\frac{1}{\gamma}+\frac{1}{\alpha}\right)+c \cdot\left(\frac{1}{\alpha}+\frac{1}{\beta}\right) \geq 2 \cdot\left(\frac{a}{\alpha}+\frac{b}{\beta}+\frac{c}{\gamma}\right)
$$