Proving an inequality involving a definite integral.

Question

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?

Thanks for any answers in advance.

Answer 1

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.

Answer 2

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 $1$.

Answer 3

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 ${1\over 4\sqrt{2}\,n^2}.$

Answer 4

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 $x\in [0,1]$.