How to prove by Mathematical Induction. I want to know how to prove this inequality by mathematical induction:
$a_k's$ are nonnegative numbers. Prove that$$a_1a_2\cdots a_n\leq \left(\frac{a_1+a_2+\cdots+a_n}{n}\right)^n.$$
In the inductive step, I tried using the inequality $ab\leq \frac{k}{k+1}a^\frac{k+1}{k}+\frac{1}{k+1}b^{k+1}$ but I got over estimate of what I was looking for.
Is there a better way of proving this by induction?
 A: Maybe this is one of those cases in which an induction proof is easier if one makes the statement to be proved stronger, because then one has a stronger induction hypothesis to use. The inequality in the question is
$$
a_1^{1/n}\cdots a_n^{1/n} \le \frac 1 n \left(a_1 + \cdots+ a_n\right).
$$
These are a geometric mean and an arithmetic mean of $n$ numbers with equal weights $1/n$.  More generally one has weights $p_1,\ldots,p_n\ge 0$, $p_1+\cdots+p_n=1$.  The inequality is then
$$
a_1^{p_1} \cdots a_n^{p_n} \le p_1 a_1 + \cdots + p_n a_n.
$$
Now let's do the induction step:
\begin{align}
a_1^{1/(n+1)} \cdots a_{n+1}^{1/(n+1)} & = \Big(a_1^{1/n} \cdots a_n^{1/n}\Big)^{n/(n+1)} \Big( a_{n+1} \Big)^{1/(n+1)} & \text{(weights $\frac n {n+1}$ and $\frac 1 {n+1}$)} \\[10pt]
& \le \frac n {n+1} \Big(a_1^{1/n} \cdots a_n^{1/n}\Big) + \frac 1 {n+1} a_{n+1} \\
& {}\qquad \text{(by the induction hypothesis in case 2)} \\[12pt]
& \le \frac n {n+1} \Big( \frac 1 n \left( a_1+\cdots+a_n \right) \Big) + \frac 1 {n+1} a_{n+1} \\
& {} \qquad \text{(by the induction hypothesis in case $n$)} \\[10pt]
& = \frac 1 {n+1} \left(a_1 + \cdots + a_{n+1} \right).
\end{align}
