How to prove the harmonic-geometric mean inequality by solving an optimization?

Question

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 $(1)$?

Answer 1

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.

Answer 2

Because by AM-GM: $$\left(\prod_{i=1}^nx_i\right)^{\frac{1}{n}}\sum_{i=1}^n\frac{1}{x_i}\geq\left(\prod_{i=1}^nx_i\right)^{\frac{1}{n}}\cdot n\left(\prod_{i=1}^n\frac{1}{x_i}\right)^{\frac{1}{n}}=\frac{n\left(\prod\limits_{i=1}^nx_i\right)^{\frac{1}{n}}}{\left(\prod\limits_{i=1}^nx_i\right)^{\frac{1}{n}}}=n.$$

Answer 3

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 $$f(x)\geq f(1)=0.$$