Hey 2021 !!

According to AM-HM ,

$\frac{2021}{a_1+a_2+\dots+a_{2021}}\leq\frac{\frac{1}{a_1}+ \frac{1}{a_2}+\dots + \frac{1}{a_{2021}} }{2021}$ ,

So,

$\frac{1}{a_1}+ \frac{1}{a_2}+\dots + \frac{1}{a_{2021}} \geq2021.$

Let's try Lagrange Multipliers here. Let $f(a_{1},...,a_{2021}) = \Sigma_{k=1}^{2021} \frac{1}{a_{k}}$ and $g(a_{1},...,a_{2021}) = \Sigma_{k=1}^{2021} a_{k} = 2021.$ Taking $\nabla f = \lambda \cdot \nabla g$ , we obtain the following:

$-\frac{1}{a^{2}_{1}} = -\frac{1}{a^{2}_{2}} = ... = -\frac{1}{a^{2}_{2021}} = \lambda \Rightarrow a_{1} = a_{2} = ... = a_{2021}.$

which after substituting into $g$ yields:

$\Sigma_{k=1}^{2021} a_{1} = 2021 \Rightarrow 2021a_{1} = 2021 \Rightarrow a_{1} = 1$ .

Taking the Hessian matrix of $f$ :

$F(a_{1}, ... , a_{2021}) = \begin{bmatrix} 2/a^{3}_{1} & 0 & \dots & 0 \\ 0 & 2/a^{3}_{2} & \dots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \dots & 2/a^{3}_{2021} \end{bmatrix}$

at the critical point $(a_{1}, ... , a_{2021}) = (1, ..., 1)$ produces $F(1, ... , 1) = 2 \cdot I_{2021 x 2021} > 0$ , which is positive-definite and hence a global minimum. The minimum value for $f$ computes to $2021 \cdot 1 = \boxed{2021}.$