Balti Tee 1991 · Ülesanne 5

To prove the first inequality, note that $\frac{2}{a+b} \leq \frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)$ and similarly $\frac{2}{b+c} \leq \frac{1}{2}\left(\frac{1}{b}+\frac{1}{c}\right), \frac{2}{c+a} \leq$ $\frac{1}{2}\left(\frac{1}{c}+\frac{1}{a}\right)$. For the second part, use the inequality $\frac{3}{x+y+z} \leq \frac{1}{3}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)$ for $x=a+b, y=b+c$ and $z=c+a$.