# Proving $\sqrt{n}+\frac{1}{\sqrt{n+1}} \geq \sqrt{n+1}$

I would like to know how to prove the following assertion :

For every $n>0$: $$\sqrt{n}+\frac{1}{\sqrt{n+1}} \geq \sqrt{n+1}$$

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$$\sqrt{n}+ \frac{1}{\sqrt{n+1}} \ge \sqrt{n+1} \\ \sqrt{n + 1}\sqrt{n} + 1 \ge n+1 \\ \sqrt{n + 1}\sqrt{n} \ge n \\ \sqrt{n + 1}\sqrt{n} \ge \sqrt{n}\sqrt{n} \\ \sqrt{n + 1}\sqrt{n} \ge \sqrt{n}\sqrt{n} \\ \sqrt{n + 1} \ge \sqrt{n} \\$$ – user2468 Aug 5 '12 at 18:04
@J.D.: Why not an answer? – joriki Aug 5 '12 at 18:07

There are many approaches. One of them is to try to prove the equivalent assertion that $\sqrt{n+1}-\sqrt{n} \le \frac{1}{\sqrt{n+1}}$.

Note that $$\sqrt{n+1}-\sqrt{n}=\frac{(\sqrt{n+1}-\sqrt{n})(\sqrt{n+1}+\sqrt{n})}{\sqrt{n+1}+\sqrt{n}}=\frac{1}{\sqrt{n+1}+\sqrt{n}}.\tag{1}$$ It is clear that for $n \gt 0$, the right-hand side of $(1)$ is $\lt \frac{1}{\sqrt{n+1}}$.

The calculation gives a stronger estimate. The right-hand side of $(1)$ is actually close to $\frac{1}{2\sqrt{n+1}}$ when $n$ is large.

Another way: Equivalently, we want to prove that $\sqrt{n+1}-\frac{1}{\sqrt{n+1}} \le \sqrt{n}$. We have $$\sqrt{n+1}-\frac{1}{\sqrt{n+1}}=\frac{(n+1)-1}{\sqrt{n+1}}=\frac{n}{\sqrt{n+1}}.$$ Since $\sqrt{n+1} \gt \sqrt{n}$, we conclude that if $n \gt 0$ then $\frac{n}{\sqrt{n+1}}\lt \sqrt{n}$.

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$$\sqrt{n}+\frac{1}{\sqrt{n+1}}\geq \sqrt{n+1}$$ $$\sqrt{n}\sqrt{n+1}+1\geq n+1$$ $$\sqrt{n}\sqrt{n+1}\geq n$$ $$n(n+1)\geq n^2$$ $$n^2+n\geq n^2$$ $$n\geq 0$$

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Or you could obfuscate the problem. Let $t=\sqrt{n+1}$. Then we have $t>1$ and $\sqrt{n}=\sqrt{t^2-1}$, so \begin{align*} &\sqrt{t^2-1}+\frac{1}{t} \ge t\\ \text{iff}\qquad &\sqrt{t^2-1}\ge t-\frac{1}{t}\\ \text{iff}\qquad &t^2-1 \ge t^2-2+\frac{1}{t^2}\\ \text{iff}\qquad &1 \ge \frac{1}{t^2}\\ \text{iff}\qquad &t^2\ \ge 1 \end{align*} which is true, since $t>1$.

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