# How to prove the following inequality: $\frac{\sqrt{n + 1}}{\sqrt{n}} - 1 \leq \frac{1}{2n - 1}$

As a part of my practice for an upcoming mid-term, I managed to simplify the following inequality to what you see here: $$\frac{\sqrt{n + 1}}{\sqrt{n}} - 1 \leq \frac{1}{2n - 1}$$

And honestly I'm stuck. If it matters, I simply need to prove this for $n \geq 1$. I also don't think we're supposed to use derivatives to prove this.

Any help is well appreciated!

EDIT: Thank you to Barry, as well as Did for the help, I've since completed the proof. That said I apologize, I hate the idea of using this website for essentially "Do these questions for me!" but I felt it was alright because these were optional rather than part of an assignment.

As a tangent, alkabary could you elaborate on how to use induction to prove this? This inequality was actually what I needed to prove to end off a part of a bigger induction proof, so I'd love to see how I could use induction to prove this smaller part, since I'm learning the fundamentals of induction right now.

• Induction man !! for all integers $n \geq 1$ – alkabary Jun 5 '15 at 20:05
• Direct approach: one wants to show that $$\frac{\sqrt{n+1}}{\sqrt{n}}\leqslant1+\frac1{2n-1}=\frac{2n}{2n-1}.$$ Since everything is positive, one can square both sides and reduce the result to a common denominator. This yields $$(n+1)(2n-1)^2\leqslant4n^3.$$ Simplify, one is left with $$3n-1\geqslant0,$$ QED. – Did Jun 5 '15 at 20:22
• Ignore alkabary's suggestion. Many people, when they see a question involving $n$ in any shape or form, will immediately think of induction; this is normal, and sometimes leads to a solution. But some people will immediately post a comment recommending induction, without trying it for themselves; this is annoying. – TonyK Jun 5 '15 at 21:05

$${\sqrt{n+1}\over\sqrt n}-1={\sqrt{n+1}-\sqrt n\over\sqrt n}={(n+1)-n\over\sqrt n(\sqrt{n+1}+\sqrt n)}={1\over\sqrt{n^2+n}+n}$$
That is equivalent to: $$\sqrt{1+\frac{1}{n}}\leq\frac{1}{1-\frac{1}{2n}}$$ that follows from the AM-GM inequality, since: $$\left(1-\frac{1}{2n}\right)^2\cdot\left(1+\frac{1}{n}\right) \leq 1.$$
• Just a note: Plugging $a=1$ and $b=1+{1\over n}$ into the AM-GM inequality gives the stronger inequality $\sqrt{1+{1\over n}}\le 1+{1\over2n}$. By stronger, I mean $1/(2n)\le1/(2n-1)$. – Barry Cipra Jun 5 '15 at 20:43