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 1


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


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