Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

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

share|cite|improve this question
$$ \sqrt{n}+ \frac{1}{\sqrt{n+1}} \ge \sqrt{n+1} \\ \sqrt{n + 1}\sqrt{n} + 1 \ge n+1 \\ \sqrt{n + 1}\sqrt{n} \ge n \\ \sqrt{n + 1}\sqrt{n} \ge \sqrt{n}\sqrt{n} \\ \sqrt{n + 1}\sqrt{n} \ge \sqrt{n}\sqrt{n} \\ \sqrt{n + 1} \ge \sqrt{n} \\ $$ – 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 $\frac{n}{\sqrt{n+1}}\lt \sqrt{n}$.

share|cite|improve this answer

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

share|cite|improve this answer

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 $t>1$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.