7
$\begingroup$

I stumbled on the following inequality: For all $n\geq 1,$ $$2\sqrt{n+1}-2\sqrt{n}<\frac{1}{\sqrt{n}}<2\sqrt{n}-2\sqrt{n-1}.$$ However I cannot find the proof of this anywhere. Any ideas how to proceed?

Edit: I posted a follow-up question about generalizations of this inequality here: Square root inequality revisited

$\endgroup$
15
$\begingroup$

$f(x)=\frac{1}{\sqrt{x}}$ is a decreasing function on $\mathbb{R}^+$, hence: $$ 2\sqrt{n+1}-2\sqrt{n}= \int_{n}^{n+1}\frac{dx}{\sqrt{x}} \leq \frac{1}{\sqrt{n}} $$ as well as $$ 2\sqrt{n}-2\sqrt{n-1}=\int_{n-1}^{n}\frac{dx}{\sqrt{x}}\geq \frac{1}{\sqrt{n}}.$$

| cite | improve this answer | |
$\endgroup$
11
$\begingroup$

$$2\sqrt{n+1}-2\sqrt{n}=2\left(\sqrt{n+1}-\sqrt{n}\right)=$$ $$=\frac2{\sqrt{n+1}+\sqrt{n}}<\frac{1}{\sqrt{n}} \Leftrightarrow$$ $$\Leftrightarrow 2\sqrt n<\sqrt{n+1}+\sqrt{n}\Leftrightarrow\sqrt n< \sqrt{n+1}$$

| cite | improve this answer | |
$\endgroup$
8
$\begingroup$

\begin{align*} 2\sqrt{n+1}-2\sqrt{n} &= 2\frac{(\sqrt{n+1}-\sqrt{n})(\sqrt{n+1}+\sqrt{n})}{(\sqrt{n+1}+\sqrt{n})} \\ &= 2\frac{1}{(\sqrt{n+1}+\sqrt{n})} \\ &< \frac{2}{2\sqrt{n}} \text{ since } \sqrt{n+1} > \sqrt{n}\\ &=\frac{1}{\sqrt{n}} \end{align*} Similar proof for the other inequality.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ I think this is a copy of Roman83 's answer. $\endgroup$ – Dragonemperor42 Jul 15 '16 at 11:06
  • $\begingroup$ I saw the answer only after I posted mine. $\endgroup$ – user348749 Jul 15 '16 at 11:20
  • 1
    $\begingroup$ 5 minutes time lag for a nearly copied answer is not accepted, even if you are on a slow network. This way someone can answer any question on this site, claiming they saw it only after their answer was posted. $\endgroup$ – Dragonemperor42 Jul 15 '16 at 11:23
  • 12
    $\begingroup$ @Roby5 - calm down. It usually takes far more than 5 minutes to compose even a short post like this one. Don't berate people because someone else has the same basic idea (very common at this level of mathematics) and happened along a few minutes earlier. Such posts are only unacceptable when the later poster clearly had time to see the other post before they started their own. $\endgroup$ – Paul Sinclair Jul 15 '16 at 13:42
5
$\begingroup$

Let $f(x)=2\sqrt{x}$. Using mean value theorem we get $$\frac{f(n+1)-f(n)}{(n+1)-n} = f'(c)$$ for some $c \in (n, n+1)$. Equivalently $$2\sqrt{n+1} - 2\sqrt{n} = \frac{1}{\sqrt c}.$$ Since $c>n$, $$\frac{1}{\sqrt c} < \frac{1}{\sqrt{n}},$$ therefore $$2\sqrt{n+1}-2\sqrt{n} < \frac 1{\sqrt n}.$$

Right inequality can be proved in a similar manner.

| cite | improve this answer | |
$\endgroup$
3
$\begingroup$

As $\sqrt n>0,$

$$2\sqrt{n+1}-2\sqrt{n}<\frac{1}{\sqrt{n}}\iff2\sqrt{n(n+1)}-2n<1$$ $$\iff2\sqrt{n(n+1)}<2n+1$$

Squaring we get $$4(n^2+n)<(2n+1)^2=4n^2+4n+1\iff 1>0$$ which is true.

Can you follow the same method for the other inequality?

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Yes, for sure. Thank you. $\endgroup$ – themightymoose Jul 15 '16 at 10:59
3
$\begingroup$

$$2\sqrt{n+1}-2\sqrt{n}<\frac{1}{\sqrt{n}}<2\sqrt{n}-2\sqrt{n-1}\iff\sqrt{n^2+n}-n\lt\frac 12\lt n-\sqrt{n^2-1}$$ Put $X=n+\frac 12$ and $Y=n-\frac 12$ so you get the evidences $$\sqrt{X^2-\frac 14}\lt X\iff X^2-\frac 14\lt X^2$$ and $$\sqrt{Y^2-\frac 14}\lt Y\iff Y^2-\frac 14\lt Y^2$$

| cite | improve this answer | |
$\endgroup$

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