Don't know if it is "the easiest", but it's certainly a pretty straightforward way of doing it. It's a little simpler to prove
$$(n+1)(a_1+\cdots+a_n) \leq n(a_1+\cdots+a_{n+1})$$
which is equivalent.
The case $n=1$ is easy (as usual): $2a_1 = a_1+a_1\leq a_1+a_2 = 1(a_1+a_2)$, so you are set.
Assuming the case $n=k$ holds, you want to establish the case $n=k+1$; you assume
$$(k+1)(a_1+\cdots+a_k) \leq k(a_1+\cdots + a_{k+1})$$
and you want to show that
$$(k+2)(a_1+\cdots+a_{k+1})\leq (k+1)(a_1+\cdots+a_{k+2}).$$
You have:
$$\begin{align*}
(k+2)(a_1+\cdots+a_{k+1}) &= (k+1)(a_1+\cdots+a_k) + (a_1+\cdots+a_k) + (k+2)a_{k+1}\\
&\leq k(a_1+\cdots+a_{k+1}) + (a_1+\cdots+a_k) + (k+2)a_{k+1}\\
&= (k+1)(a_1+\cdots+a_k) + ka_{k+1} + (k+2)a_{k+1}\\
&= (k+1)(a_1+\cdots +a_k) + (k+1)a_{k+1} + (k+1)a_{k+1}\\
&= (k+1)(a_1+\cdots+a_{k+1}) + (k+1)a_{k+1}\\
&\leq (k+1)(a_1+\cdots+a_{k+1}) + (k+1)(a_{k+2})\\
&= (k+1)(a_1+\cdots+a_{k+2}),
\end{align*}$$
Which proves the induction.