Hey $2019!!$

According to AM-HM ,

$\frac{2019}{a_1+a_2+\dots+a_{2019}}\leq\frac{\frac{1}{a_1}+ \frac{1}{a_2}+\dots + \frac{1}{a_{2019}} }{2019}$ ,

So,

$\frac{1}{a_1}+ \frac{1}{a_2}+\dots + \frac{1}{a_{2019}} \geq2019.$

By Titu's lemma :

$\begin{aligned} \frac {1^2}{a_1} + \frac {1^2}{a_2} + \frac {1^2}{a_3} + \cdots + \frac {1^2}{a_{2019}} \ge \frac {(1+1+1+\cdots + 1)^2}{a_1+a_2+a_3+\cdots + a_{2019}} = \frac {2019^2}{2019} = \boxed{2019} \end{aligned}$

Equality occurs when $a_1=a_2=a_3=\cdots=a_{2019} = 1$ .

With $x_j = \sqrt{a_j}$ and $y_j = \tfrac{1}{\sqrt{x_j}}$ for $1 \le j \le 2019$ , the Cauchy-Schwarz inequality gives $2019^2 \; = \; \left(\sum_{j=1}^{2019}x_jy_j\right)^2 \; \le \; \left(\sum_{j=1}^{2019}a_j\right)\left(\sum_{j=1}^{2019}a_j^{-1}\right) \; = \; 2019\left(\sum_{j=1}^{2019}a_j^{-1}\right)$ so that $\sum_{j=1}^{2019}a_j^{-1} \;\ge \; \boxed{2019}$ with equality achieved when $a_1 = a_2 = \cdots = a_{2019} = 1$ .

This can be solved in many ways, including Cauchy Schwarz, AM-GM, AM-HM, etc. I will present a solution using Cauchy Schwarz.

Applying Cauchy, we have $\begin{aligned}(a_1+a_2+\cdots+a_{2019})\left(\frac{1}{a_1}+\frac 1{a_2}+\cdots + \frac 1{a_{2019}}\right) &\ge \left(\frac{\sqrt{a_1}}{\sqrt{a_1}} + \frac{\sqrt{a_2}}{\sqrt{a_2}} + \cdots +\frac{\sqrt{a_{2019}}}{\sqrt{a_{2019}}}\right)^2 \\ &= (\underbrace{1+1+\cdots + 1}_{2019})^2 \\ &= 2019^2. \end{aligned}$ Noting that $a_1+a_2+\cdots+a_{2019} = 2019$ , we have $\frac{1}{a_1}+\frac 1{a_2}+\cdots + \frac 1{a_{2019}} \ge 2019,$ which means that the minimum value is $\boxed{2019}$ .

If you want the minimum value then $a1, a2, a3,...,a2019$ are all equal, therefore they can only be equal to 1, so the answer is 2019!

a1+a2+a3.......a2019=2019 that will happen iff all as from a1 to a2019 equal 1 so we just add up ones 2019 times which equals 2019