Pierādīt nevienādību
\[\frac{1}{\sqrt{1}+\sqrt{3}}+\frac{1}{\sqrt{2}+\sqrt{4}}+\frac{1}{\sqrt{3}+\sqrt{5}}+\frac{1}{\sqrt{4}+\sqrt{6}}+\cdots+\frac{1}{\sqrt{48}+\sqrt{50}} < 6\]
Ievērojam, ka \(\frac{1}{\sqrt{n}+\sqrt{n+2}}=\frac{\sqrt{n+2}-\sqrt{n}}{(\sqrt{n+2}+\sqrt{n})(\sqrt{n+2}-\sqrt{n})}=\frac{\sqrt{n+2}-\sqrt{n}}{n+2-n}=\frac{\sqrt{n+2}-\sqrt{n}}{2}\).
Pārveidojot dotās nevienādības kreiso pusi, iegūstam
\[\begin{aligned} & \frac{1}{\sqrt{1}+\sqrt{3}}+\frac{1}{\sqrt{2}+\sqrt{4}}+\frac{1}{\sqrt{3}+\sqrt{5}}+\frac{1}{\sqrt{4}+\sqrt{6}}+\cdots+\frac{1}{\sqrt{48}+\sqrt{50}}= \\ & =\frac{\sqrt{3}-\sqrt{1}+\sqrt{4}-\sqrt{2}+\sqrt{5}-\sqrt{3}+\sqrt{6}-\sqrt{4}+\cdots+\sqrt{49}-\sqrt{47}+\sqrt{50}-\sqrt{48}}{2}= \\ & =\frac{-\sqrt{1}-\sqrt{2}+\sqrt{49}+\sqrt{50}}{2}=3+2 \sqrt{2}<3+2 \cdot 1,5=6 \end{aligned}\]