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?

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}
