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Với mọi $x$ dương ta luôn có:
$x+(x+1)\geq 2\sqrt{x(x+1)}$ (BĐT Cô si)
Do dấu bằng ở không xảy ra nên ta luôn có:
$x+(x+1)> 2\sqrt{x(x+1)}$
$\Leftrightarrow \frac{1}{x+(x+1)}<\frac{1}{2\sqrt{x(x+1)}}$
$\Leftrightarrow \frac{\sqrt{x+1}-\sqrt{x}}{x+(x+1)}<\frac{\sqrt{x+1}-\sqrt{x}}{2\sqrt{x(x+1)}}=\frac{1}{2\sqrt{x}}-\frac{1}{2\sqrt{x+1}}$
(Do $\sqrt{x+1}-\sqrt{x}>0$)
Áp dụng điều trên:
$\frac{\sqrt{2}-\sqrt{1}}{1+2}<\frac{1}{2}-\frac{1}{2\sqrt{2}}$
$\frac{\sqrt{3}-\sqrt{2}}{2+3}<\frac{1}{2\sqrt{2}}-\frac{1}{2\sqrt{3}}$
$\frac{\sqrt{4}-\sqrt{3}}{3+4}<\frac{1}{2\sqrt{3}}-\frac{1}{2\sqrt{4}}$
.....
$\frac{\sqrt{25}-\sqrt{24}}{24+25}<\frac{1}{10}-\frac{1}{2\sqrt{24}}$
Cộng theo vế :
$\frac{\sqrt{2}-\sqrt{1}}{1+2}+\frac{\sqrt{3}-\sqrt{2}}{2+3}+\frac{\sqrt{4}-\sqrt{3}}{3+4}+...\frac{\sqrt{25}-\sqrt{24}}{24+25}<\frac{1}{2}-\frac{1}{10}=\frac{2}{5}$
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