How to prove the harmonic-geometric mean inequality by solving an optimization? The harmonic-geometric mean inequality is defined as follows
$$
\frac{n}{\sum_{i=1}^n \frac{1}{x_i}} \leq (\Pi_{i=1}^{n}x_i)^{\frac{1}{n}}\tag{1}
$$
Given the following linear programming problem
$$
\min \sum_{i=1}^n \frac{1}{x_i}\\
\begin{align}
\text{s.t} \,\,\,\,\,\,\,& \Pi_{i=1}^{n}x_i=1\\
&x\geq0
\end{align}
$$
where $x \in \mathbb{R}^n$.
If we set up KKT conditions, we end up with $x  =[1, \cdots, 1]^{\top}$ as the optimal point of the optimization. Hence the minimum value is $n$.
Question: using the above result how we can prove $(1)$?
 A: Suppose we have $y_1, \ldots, y_n > 0.$ Define $P = \prod_{i=1}^n y_i$ and $x_i = y_i  \cdot P^{-1/n}.$ Then $x_i \geq 0$ and $\prod_{i=1}^n x_i = 1$ so by the result of the optimization problem we have 
$$ \sum_{i=1}^n \frac{1}{x_i} \geq n$$
Since $x_i = y_i \cdot P^{-1/n}$ we have 
$$\sum_{i=1}^n \frac{P^{1/n}}{y_i} \geq n$$
which rearranges to $$\frac{n}{\sum_{i=1}^n \frac{1}{y_i}} \leq P^{1/n}$$ as required. 
A: Because by AM-GM:
$$\left(\prod_{i=1}^nx_i\right)^{\frac{1}{n}}\sum_{i=1}^n\frac{1}{x_i}\geq\left(\prod_{i=1}^nx_i\right)^{\frac{1}{n}}\cdot n\left(\prod_{i=1}^n\frac{1}{x_i}\right)^{\frac{1}{n}}=\frac{n\left(\prod\limits_{i=1}^nx_i\right)^{\frac{1}{n}}}{\left(\prod\limits_{i=1}^nx_i\right)^{\frac{1}{n}}}=n.$$
A: We can use also the TL method:
Since our inequality is homogeneous, we can assume  $\prod\limits_{i=1}^nx_i=1.$
Thus, we need to prove that:
$$\sum_{i=1}^n\frac{1}{x_i}\geq n.$$
Indeed, $$\sum_{i=1}^n\frac{1}{x_i}-n=\sum_{i=1}^n\left(\frac{1}{x_i}-1+\ln{x_i}\right)\geq0$$
because easy to show that for any positive $x$ the following inequality is true.
$$\frac{1}{x}-1+\ln{x}\geq0.$$
Indeed, let $f(x)=\frac{1}{x}-1+\ln{x}.$
Thus, $$f'(x)=-\frac{1}{x^2}+\frac{1}{x}=\frac{x-1}{x^2},$$ which says that $x_{min}=1$ and
$$f(x)\geq f(1)=0.$$
