How to prove $(\frac 1n)^n+(\frac 2n)^n+\cdots+(\frac nn)^n\geqslant\frac{3n+1}{2n+2}$

How to prove

$$\Bigl(\dfrac 1n\Bigr)^n + \Bigl(\frac 2n\Bigr)^n + \cdots + \Bigl(\frac nn\Bigr)^n \geqslant \frac{3n+1}{2n+2} \qquad (n\in\mathbb{N})$$

I tried:

let $f(x)=x^n$ and $f''(x)\geqslant 0$ for $n>1$, let $x_i=\frac in$, and we have

$$\sum f(x_i)\geqslant nf\biggl(\frac{\sum x_i}{n}\biggr).$$

But the $RHS<\dfrac{3n+1}{2n+2}$. So it doesn't work.

Could someone give me a neat proof? Thanks!

1 Answer

Using trapezoidal integration approximation to $\int_0^1x^ndx$ with step $\frac1n$ $$\Bigl(\frac 1n\Bigr)^n + \Bigl(\frac 2n\Bigr)^n + \cdots + \Bigl(\frac nn\Bigr)^n>n\int_0^1x^ndx+\frac12\Bigl(\frac nn\Bigr)^n=\frac{3n+1}{2n+2}$$