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Ta có
$\frac{1}{\sqrt{k}}=\frac{2}{2\sqrt{k}}>\frac{2}{\sqrt{k}+\sqrt{k+1}}=\frac{2\left (\sqrt{k+1} - \sqrt{k}\right )}{k+1-k}=2\left (\sqrt{k+1} - \sqrt{k}\right )$
Suy ra
$\frac{1}{\sqrt{1}}>2\left (\sqrt{2} - \sqrt{1}\right )$
$\frac{1}{\sqrt{2}}>2\left (\sqrt{3} - \sqrt{2}\right )$
$...$
$\frac{1}{\sqrt{2005}}>2\left (\sqrt{2006} - \sqrt{2005}\right )$
cộng theo từng vế các BDT này ta có đpcm.
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