Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

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

(how) Can I show that:

$n \ge \sqrt{n+1}+\sqrt{n}$ ?

It should be true for all $n \ge 5$.

Tried it via induction:

  1. $n=5$: $5 \ge \sqrt{5} + \sqrt{6} $ is true.

  2. $n\implies n+1$: I need to show that $n+1 \ge \sqrt{n+1} + \sqrt{n+2}$

Starting with $n+1 \ge \sqrt{n} + \sqrt{n+1} + 1 $ .. (now??)

Is this the right way?

share|cite|improve this question
You then would like to show that $\sqrt{n} + \sqrt{n+1} + 1 \geq \sqrt{n+1} + \sqrt{n+2}$, that is, $\sqrt{n} + 1 \geq \sqrt{n+2}$. Try squaring both sides. – TMM May 26 '13 at 17:51

Hint: $\sqrt{n} + \sqrt{n+1} \leq 2\sqrt{n+1}$. Can you take it from there?

share|cite|improve this answer

Here is another way:

Define $f(x)=x-\sqrt{x}-\sqrt{x+1}$. We need to show that $f(x)\ge 0$ for all $x\ge 5$. Since $$f'(x)=1-\frac{1}{2\sqrt{x}}-\frac{1}{2\sqrt{x+1}} \ge 0, \quad \forall x\ge 1$$ the function is increasing on $[1,\infty)$. As $f(5)\ge 0$, the result follows.

share|cite|improve this answer

What I would try first if I wanted it to be true - no tricks:

$n \ge \sqrt{n + 1} + \sqrt{n} \iff \frac{n}{\sqrt{n}} \ge \frac{\sqrt{n+1}}{\sqrt{n}} + \frac{\sqrt{n}}{\sqrt{n}} \iff \sqrt{n} \ge \sqrt{1 + \frac{1}{n}} + 1 \Leftarrow n \ge 5$.

share|cite|improve this answer

To add to your step, observe the following:

$$\sqrt{n}+1 = \sqrt{(\sqrt{n}+1)^2} = \sqrt{n+1+2\sqrt{n}} > \sqrt{n+1+1} = \sqrt{n+2}.$$

The "$>$" part comes from your assumption $n \ge 5$, so $2\sqrt{n} \ge 2\sqrt{5} > 1$. Now, we have:

$$n+1 \ge \sqrt{n} + \sqrt{n+1} + 1 = \sqrt{n+1} + (\sqrt{n}+1) > \sqrt{n+1} + \sqrt{n+2}.$$

share|cite|improve this answer

First observe that since $5 \le n$, we have: $$ 4(n+1) =4n+4 < 4n+5 \le 4n+n = 5n \le (n)n = n^2 $$ Hence, since $4(n+1)<n^2 \Rightarrow \boxed{n+1<\dfrac{1}{4}n^2}$ (and $f(x)=\sqrt{x}$ is monotonically increasing and $n\ge0$), we have:

$$ \sqrt{n+1}+\sqrt{n} \le \sqrt{n+1}+\sqrt{n+1} = 2\sqrt{n+1}<2\sqrt{\dfrac{1}{4}n^2}=2\left(\dfrac{1}{2}n\right) = n $$

as desired.

share|cite|improve this answer

You can replace the $\sqrt{n}$ on the RHS with another $\sqrt{n+1}$. Therefore you have $n\ge2\sqrt{n+1}$, or $n^2\ge4n+4$. $n^2-4n+4\ge8$, or $(n-2)^2\ge8$. The lowest integer solution to this is $5$, so $n\ge5.$

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.