Balti Tee 1994 · Ülesanne 14

Solution: Clearly, the inequality $a > b$ implies $\alpha > \beta$ and similarly $a < b$ implies $\alpha < \beta$, hence $(a-b)(\alpha-\beta) \geq 0$ and $a \alpha + b \beta \geq a \beta + b \alpha$. Dividing the last equality by $\alpha \beta$ we get

$$\frac{a}{\beta} + \frac{b}{\alpha} \geq \frac{a}{\alpha} + \frac{b}{\beta}$$

Similarly we get

$$\frac{a}{\gamma} + \frac{c}{\alpha} \geq \frac{a}{\alpha} + \frac{c}{\gamma}$$

and

$$\frac{b}{\gamma} + \frac{c}{\beta} \geq \frac{b}{\beta} + \frac{c}{\gamma}$$

To finish the proof it suffices to add the inequalities (6)-(8).