Power Mean Inequalities

Simplifying the expression $(a_1+a_2+a_3+........+a_n) + 4 ( \dfrac{1}{a_!}+ \dfrac{1}{a_2} + \dfrac{1}{a_3}+.......+\dfrac{1}{a_n}) \geq 2017 \\ \\ => \sum_{i=1}^{n} (a_i +\dfrac{4}{a_i}) \geq 2017$

Now applying A.M.-G.M. $a_i+ \dfrac{4}{a_i} \geq 2 \sqrt{a_i \times \dfrac{4}{a_i}}=4$

We will find the minimum value of $n$ .The $n$ will minimum when it will holds the equality.So each $a_i+ \dfrac{4}{a_i}$ will be $4$

So, putting this value $\sum_{i=1}^{n} 4 \geq 2017 \implies 4n \geq 2017 \implies n \geq 2017/4=504.25$

The next integer $n=505$ will be the minimum value.