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The problem asks to prove the inequality:

\[
\frac{1}{\sqrt{n+1} + \sqrt{n}} < \frac{1}{2\sqrt{n}}
\]

for any natural number \( n \).

### Solution:

Let's start by simplifying the left-hand side of the inequality:

1. **Rationalizing the denominator of** \( \frac{1}{\sqrt{n+1} + \sqrt{n}} \):

Multiply both the numerator and denominator by \( \sqrt{n+1} - \sqrt{n} \):

\[
\frac{1}{\sqrt{n+1} +

Montrer que \frac{1}{\sqrt{n+1} + \sqrt{n}} < \frac{1}{2\sqrt{n}} pour n entier naturel non nul.

Explore how to rationalize and prove the inequality \frac{1}{\sqrt{n+1} + \sqrt{n}} < \frac{1}{2\sqrt{n}}. This involves rationalizing the denominator and comparing expressions for a clear understanding of radicals and inequalities.

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