Prove $\int_{0}^{1} \sqrt{x^n+1 }dx \geq \frac{2}{n+2}$ for any $n \in \mathbb{N}$

Question

Prove $\int_{0}^{1} \sqrt{x^n+1 }dx \geq \frac{2}{n+2}$ for any $n \in \mathbb{N}$

I've tried to do it with monotony, but I only got that the integral is bigger than $1$. However, $n$ is a natural number (except $0$), so I need to prove that the integral is bigger than $2/3$ at least.

Thanks for the help

Answer 1

$\int_0^{1} \sqrt {x^{n}+1} dx >\int_0^{1} {x^{n/2}} dx=\frac 1 {\frac n 2 +1}=\frac 2 {n+2}$

Answer 2

$$\displaystyle{\int_{0}^{1}}\sqrt{x^n+1}dx \ge \int_{0}^{1}1dx =1 \ge \dfrac{2}{2+n} \\ \text{for all } n \in \mathbb{N}$$

Answer 3

By AM-GM $$\int\limits_{0}^{1} \sqrt{x^n+1 }dx \geq\int\limits_{0}^{1} \sqrt{2\sqrt{x^n} }dx=\frac{\sqrt2}{\frac{n}{4}+1}>\frac{2}{n+2}.$$