Prove the following inequality by an induction argument In some note I see  the writer claims that the following inequality is obvious  using an  induction argument:
\begin{equation}
\left(\left(\frac{1}{n}\right)^n+\left(\frac{2}{n}\right)^n+\cdots+\left(\frac{n}{n}\right)^n\right)^{\frac{1}{n}}<1+\frac{1}{n}
\end{equation}
But it's not obvious to me. Could you please help me to understand the claimed obvious solution?
 A: Since my comment turned out to be an answer, I'll post it as such.
Raise both side to the $n$ and clear denominators.  We must show $$\sum_{k=1}^n{k^n}\le n^n\left(1+\frac1n\right)^n=(n+1)^n$$  However,
$$\sum_{k=1}^n{k^n}\le\int_1^{n+1}{x^n\mathrm{dx}}=\frac{x^{n+1}}{n+1}\Big|_1^{n+1}<\frac{(n+1)^{n+1}}{n+1}=(n+1)^n$$
A: I can't prove it by induction,
but I can prove it.
We want to show that
$\Big(\sum_{k=1}^n(k/n)^n\Big)^{1/n‎} 
< 1+‎\dfrac{1}{n}‎
$.
If $f(x) = x^n$,
since
$f'(x) > 0$
for $n \ge 2$ and
$0 < x < 1$,
$(k/n)^n
\lt n\int_{k/n}^{(k+1)/n} x^n dx
\lt ((k+1)/n)^n
$.
Therefore
$\begin{array}\\
\sum_{k=1}^{n-1}(k/n)^n
&=\sum_{k=0}^{n-1}(k/n)^n\\
&\lt \sum_{k=0}^{n-1}n\int_{k/n}^{(k+1)/n} x^n dx\\
&= n\sum_{k=0}^{n-1}\int_{k/n}^{(k+1)/n} x^n dx\\
&= n\int_{0}^{1} x^n dx\\
&= \dfrac{n}{n+1}\\
\text{so}\\
\sum_{k=1}^{n}(k/n)^k
&\lt 1+\frac{n}{n+1}\\
\end{array}
$
Taking the $n$-th root
$(\sum_{k=1}^{n}(k/n)^n)^{1/n}
\lt (1+\dfrac{n}{n+1})^{1/n}
$.
By Bernoulli's inequality,
$(1+x)^n \ge 1+nx$
or
$(1+x/n)^n \ge 1+x$.
Taking the $n$-th root,
$(1+x)^{1/n} \le 1+x/n$.
Putting $x=n/(n+1)$,
$(1+n/(n+1))^{1/n} 
\le 1+(n/(n+1))/n
=1+1/(n+1)
$.
Therefore
$\Big(\sum_{k=1}^n(k/n)^n\Big)^{1/n‎} 
< 1+‎\dfrac{1}{n+1}‎
$
which is slightly better
that what was asked.
We can accurately estimate the sum
by noting that
$(1-k/n)^n
\approx e^{-k}
$
for small $k$,
so that
$\begin{array}\\
\sum_{k=1}^n(k/n)^n
&=\sum_{k=0}^{n-1}((n-k)/n)^n\\
&=\sum_{k=0}^{n-1}(1-k/n)^n\\
&\approx\sum_{k=0}^{n-1}e^{-k}\\
&\approx \dfrac{1}{1-1/e}\\
&= \dfrac{e}{e-1}\\
&=1+ \dfrac{1}{e-1}\\
&\approx 1.582\\
\end{array}
$
Therefore
$(\sum_{k=1}^n(k/n)^n)^{1/n}
\approx (1+ \dfrac{1}{e-1})^{1/n}
\approx 1+ \dfrac{1}{n(e-1)}
$.
