# Proving an inequality involving a definite integral.

I have to say in advance that this question is claimed to be a “very easy” question, and that is why I am struggling with it.

Show that $$\frac{1}{(n+1)\sqrt2}\leq \int_0^1\frac{x^n}{\sqrt{1+x}}\ dx\leq \frac{1}{n+1}.$$

My approach:

It suffices to prove that $$\frac{1}{\sqrt2}\leq \int_0^1\frac{(n+1)x^n}{\sqrt{1+x}}\ dx\leq1$$.

The middle integral is equal to $$\frac{1}{\sqrt2}+\frac{1}{2}\int_0^1\frac{x^{n+1}}{(1+x)^{\frac{3}{2}}}\ dx$$, where $$\frac{x^{n+1}}{(1+x)^{\frac{3}{2}}}$$ is monotonic increasing in $$(0,1)$$ (which can be proved by differentiating and see that all the extremums are attained when $$x\leq0$$). It follows from the mean value theorem for integral that the integral takes values between $$0$$ and $$\frac{5}{8}\sqrt2$$ (which is less than 1). Therefore, the originally inequality can be proved.

My problem:

I have omitted a lot of (clumsy) detail here. Doing some research on the Internet, I found there is a recurrence for this integral (and methods like induction can be used), and there is even a general formula (which allows us to compute directly). However, the common problem of these approaches is their complexities. Is there any tricky but “very easy” method that I missed?

Note that for all $$x\in \left[0,1\right]$$ and $$n\in \mathbb{N}$$, we have that $$\frac{x^{n}}{\sqrt{2}}\leq\frac{x^{n}}{\sqrt{x+1}}\leq x^{n}.$$ Integrating, we get that $$\frac{1}{(n+1)\sqrt{2}}=\int_0^1{\frac{x^{n}}{\sqrt{2}}}dx\leq\int_0^1{\frac{x^{n}}{\sqrt{x+1}}}dx\leq \int_0^1{x^{n}}dx=\frac{1}{n+1},$$ as desired.

There is some method even more easy: $$\int_0^1 \dfrac{(n+1)x^n}{\sqrt{x+1}}\mathrm{d}x= \int_0^1 \dfrac{1}{\sqrt{1+x}}\mathrm{d}x^{n+1}=\int_0^1 \dfrac{1}{\sqrt{1+x^{\frac{1}{n+1}}}}\mathrm{d}x.$$

Noticing that $$\dfrac{1}{\sqrt{1+x^{\frac{1}{n+1}}}}$$ is montone on $$n$$, so we can just calculate the case $$n=0$$ and $$n=\infty$$. That is simple.

$$n=0$$: $$\int_0^1 \dfrac{1}{\sqrt{1+x}}\mathrm{d}x=2(\sqrt{2}-1)> \dfrac{1}{\sqrt{2}}.$$

$$n\to\infty$$: The result is $$1$$.

It is possible to improve the upper bound substantially. We have $$x+1\ge 2\sqrt{x}.$$ Hence $${x^n\over \sqrt{1+x}}\le {1\over \sqrt{2}}\,x^{n-1/4}.$$ Thus $$\int\limits_0^1{x^n\over \sqrt{1+x}}\,dx\le {1\over \sqrt{2}}{1\over n+{3\over 4}}$$ In this way the gap between the upper and the lower bound is less than $${1\over 4\sqrt{2}\,n^2}.$$

I think the easy way (and the one intended) is to bound the nonnegative function $$\frac{x^n}{\sqrt{1+x}}$$ and use the monotonicity of the integral. For example, for the right-hand side inequality use that $$\sqrt{1+x}\ge 1$$ for $$x\in [0,1]$$.