Balti Tee 1995 · Ülesanne 6

Solution: The inequality between the arithmetic and harmonic mean gives

$$\begin{aligned} & \frac{a+c}{a+b}+\frac{c+a}{c+d} \geq \frac{4}{\frac{a+b}{a+c}+\frac{c+d}{c+a}} = 4 \cdot \frac{a+c}{a+b+c+d} \\ & \frac{b+d}{b+c}+\frac{d+b}{d+a} \geq \frac{4}{\frac{b+c}{b+d}+\frac{d+a}{d+b}} = 4 \cdot \frac{b+d}{a+b+c+d} \end{aligned}$$

and adding these inequalities yields the required inequality.