The best constant is $c = 4$. We first show that, if $n$ is a positive integer and $x_1, \dots, x_n$ are positive real numbers, then
$$\sum_{k=1}^{n} \left( \frac{1}{k} \sum_{j=1}^{k} x_j \right)^2 + \frac{2}{n} \left( \sum_{k=1}^{n} x_k \right)^2 < 4 \sum_{k=1}^{n} x_k^2,$$
so $c \le 4$. To prove the above inequality, proceed by induction on $n$. The base case, $n=1$, is clear. For the induction step, let $\bar{x}_k = (x_1 + \dots + x_k)/k$, $k \ge 1$, and notice that it is sufficient to show that $(2n+3)\bar{x}_{n+1}^2 - 2n\bar{x}_n^2 < 4x_{n+1}^2$. Since $x_{n+1} = (n+1)\bar{x}_{n+1} - n\bar{x}_n$, this is equivalent to $2n(2n+1)\bar{x}_n^2 - 8n(n+1)\bar{x}_n\bar{x}_{n+1} + (4n^2+6n+1)\bar{x}_{n+1}^2 > 0$. The left-hand member is a quadratic form in $\bar{x}_n$ and $\bar{x}_{n+1}$ whose discriminant is $-2n$ and the inequality follows.

To show $c \ge 4$, we prove that
$$\sum_{k=1}^{n} \left( \frac{1}{k} \sum_{j=1}^{k} \frac{1}{\sqrt{j}} \right)^2 > 4 \sum_{k=1}^{n} \frac{1}{k} - 24.$$
Divergence of the harmonic series settles the case. Write $1/\sqrt{j} > 2(\sqrt{j+1} - \sqrt{j})$, to obtain
$$\left( \frac{1}{k} \sum_{j=1}^{k} \frac{1}{\sqrt{j}} \right)^2 > \frac{4}{k^2} \left( \sqrt{k+1} - 1 \right)^2 > \frac{4}{k} \left( 1 - \frac{2}{\sqrt{k}} \right) = \frac{4}{k} - \frac{8}{k\sqrt{k}}.$$
Finally, notice that $1/(2k\sqrt{k}) < 1/\sqrt{k-1} - 1/\sqrt{k}$, $k \ge 2$, to get
$$\sum_{k=1}^{n} \frac{1}{k\sqrt{k}} \le 3 - \frac{2}{\sqrt{n}} < 3,$$
and deduce thereby the desired inequality.