Show $(a_1+···+a_n)^2 ≤ n(a^2_1+···+a^2_n).$ [duplicate]

Want to show that $(\sum_{k=1}^n a_k)^2 \le n\sum_{k=1}^n a_k^2 $ or $s_1(n)^2 \le n s_2(n) $ where $s_m(n) =\sum_{k=1}^n a_k^m $.

True for $n=1$ (trivial) and $n=2$ (easy).

If true for $n$, then

$\begin{array}\\ s_1^2(n+1) &=(s_1(n)+a_{n+1})^2\\ &=s_1^2(n)+2a_{n+1}s_1(n)+a_{n+1}^2\\ &\le ns_2(n)+2a_{n+1}\sqrt{n s_2(n)}+a_{n+1}^2\\ \end{array} $

so we want

$\begin{array}\\ ns_2(n)+2a_{n+1}\sqrt{n s_2(n)}+a_{n+1}^2 &\le (n+1)s_2(n+1)\\ &= (n+1)(s_2(n)+a_{n+1}^2)\\ &= (n+1)s_2(n)+(n+1)a_{n+1}^2\\ \end{array} $

or $2a_{n+1}\sqrt{n s_2(n)} \le s_2(n)+na_{n+1}^2 $

or

$\begin{array}\\ 0 &\le na_{n+1}^2-2a_{n+1}\sqrt{n s_2(n)}+s_2(n)\\ &=n\left(a_{n+1}^2-2a_{n+1}\sqrt{ \dfrac{s_2(n)}{n}}+\dfrac{s_2(n)}{n}\right)\\ &=n\left(a_{n+1}-\sqrt{ \dfrac{s_2(n)}{n}}\right)^2\\ \end{array} $

which is true with equality iff $a_{n+1}=\sqrt{ \dfrac{s_2(n)}{n}} $.