The power mean inequality, with a twist!

Note that we can multiply all the terms of the original inequality by any positive constant $k$ making the sequence into $ka_1, ka_2, ..., ka_{2017}$ and obtain the original inequality after substitution and dividing out both sides by $k^2$ . Also, this inequality holds for all $n$ when $a_1 = a_2$ , so assume WLOG that $a_1 > a_2$ . This means that we can choose an arbitrary $k$ for which we obtain the sequence $b_1, b_2, b_3, ..., b_{2017}$ with $b_1 - b_2 = 1$ . The original inequality then becomes

$\large \frac{b_1^2 + b_2^2 + ... + b_{2017}^2}{2017} \geq (\frac{b_1 + b_2 + ... + b_{2017}}{2017})^2 + \frac{1}{n}$

Making the substitution $b_1 = c + 0.5, b_2 = c - 0.5$ , we get the inequality

$\large \frac{(c+0.5)^2 + (c-0.5)^2 + b_3 ^2+... + b_{2017}^2}{2017} = \frac{c^2 + c + 0.25 +c^2 - c + 0.25+b_3^2 + ... + b_{2017}^2}{2017}\geq (\frac{c+0.5 + c - 0.5 + b_3 +... + b_{2017}}{2017})^2 + \frac{1}{n}$

Which is equal to saying that

$\large \frac{c^2 + c^2 + b_3 +... + b_{2017}^2}{2017} + \frac{1}{2*2017} \geq (\frac{c+c + b_3 +... + b_{2017}}{2017})^2 + \frac{1}{n}$

Or that

$\large \frac{c^2 + c^2 + b_3 +... + b_{2017}^2}{2017} - (\frac{c+c + b_3 +... + b_{2017}}{2017})^2 \geq \frac{1}{n}- \frac{1}{2*2017}$

Now, since the LHS must be nonnegative, we can safely say that

$0 \geq \frac{1}{n}- \frac{1}{2*2017} \implies \frac{1}{4034} \geq \frac{1}{n} \implies n \geq 4034$

And so, the minimum value of $n$ is $\boxed{4034}$ .