Why is ${\sum_{i=1}^n a_i^{n+1}}\bigm/{\sum_{i=1}^{n}a_i^n} \geq \frac 1n{\sum_{i=1}^n a_i}$ Why is $$\frac{\sum_{i=1}^n a_i^{n+1}}{\sum_{i=1}^{n}a_i^n} \geq \frac{\sum_{i=1}^n a_i}{n}$$
where $n$ is some positive natural number, and all $a_i$s are assumed to be positive real number?
 A: This is the special case $(r,s) = (1,n)$ of the following inequality:
Lemma.
Suppose $r$, $s$, and $a_i$ ($1 \leq i \leq n$) are positive reals.  Then
$$
n \sum_{i=1}^n a_i^{r+s} \geq \sum_{i=1}^n a_i^r \sum_{i=1}^n a_i^s,
$$
with equality iff the $a_i$ are all equal.
The relation between the number $n$ of variables and the exponents $1,n$
is a red herring.
Proof of Lemma:
Without loss of generality $r \leq s$.  Let $x_i = a_i^s$,
and divide both sides by $n^2$, so the desired inequality becomes
$$
\left( \frac1n \sum_{i=1}^n x_i^{p+1} \right)
\geq
\left( \frac1n \sum_{i=1}^n x_i^p \right)
\left( \frac1n \sum_{i=1}^n x_i \right)
$$
where $p = r/s \leq 1$.
But the function $x \mapsto x^{p+1}$ is strictly convex upwards, so
$$
\left( \frac1n \sum_{i=1}^n x_i^{p+1} \right)
\geq
\left( \frac1n \sum_{i=1}^n x_i \right)^{p+1}
$$
with equality iff the $x_i$ are all equal; while
$x \mapsto x^p$ is convex downwards, so
$$
\left( \frac1n \sum_{i=1}^n x_i^p \right)
\leq
\left( \frac1n \sum_{i=1}^n x_i \right)^p
$$
with equality if the $x_i$ are all equal.
The desired result follows.$\diamondsuit$
A: Hint: compare 
$$ \sum_{i=1}^n a_i^n\cdot \frac{\sum_{i=1}^1a_i}{n}$$
with 
$$ \sum_{i=1}^n a_i^n\cdot a_i$$
