Lower bound for sum of reciprocals of positive real numbers

Question

I am reading an article where the author seems to use a known relationship between the sum of a finite sequence of real positive numbers $a_1 +a_2 +... +a_n = m$ and the sum of their reciprocals. In particular, I suspect that \begin{equation} \sum_{i=1}^n \frac{1}{a_i} \geq \frac{n^2}{m} \end{equation}
with equality when $a_i = \frac{m}{n} \forall i$. Are there any references or known theorems where this inequality is proven?

This interesting answer provides a different lower bound. However, I am doing some experimental evaluations where the bound is working perfectly (varying $n$ and using $10^7$ uniformly distributed random numbers).

Answer 1

by Cauchy schwarz inequality$$\left(\sum_{i=1}^{n}{\sqrt{a_i}}^2\right)\left(\sum_{i=1}^{n}\frac{1}{{\sqrt{a_i}}^2}\right)\ge (\sum_{i=1}^{n} 1)^2=n^2$$

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WLOG $a_1\ge a_2...\ge a_n$ then $\frac{1}{a_1}\le \frac{1}{a_2}..\le \frac{1}{a_n}$

So by chebyshev's inequality $$n^2=n\left(a_1\frac{1}{a_1}+a_2\frac{1}{a_2}+\cdots + a_n\frac{1}{a_n}\right)\le \left(\sum_{i=1}^{n}a_i\right)\left(\sum_{i=1}^{n}\frac{1}{a_i}\right)$$ Since $\sum_{i=1}^n a_i=m$ this implies that $$\sum_{i=1}^n \frac{1}{a_i} \ge \frac{n^2}{m}$$

Answer 2

The author is using the Arithmetic Mean - Harmonic Mean ("AM-HM") inequality: https://en.wikipedia.org/wiki/Harmonic_mean#Relationship_with_other_means

This is a popular inequality in the math olympiad world; you can find a proof here: https://artofproblemsolving.com/wiki/index.php/Root-Mean_Square-Arithmetic_Mean-Geometric_Mean-Harmonic_mean_Inequality

Answer 3

This follows form Lagrange Multipliers.

Let $S_1=\sum_{i=0}^n a_i=m$. Minimize $S_2=\sum_{i=0}^n\frac{1}{a_i}$.

Treat each $a_i$ as an independent variable.

$\lambda\frac{\partial S_1}{\partial a_i}=\frac{\partial S_2}{\partial a_i}\implies \lambda=\frac{-1}{a_i^2}\implies\forall m,n, a_m^2=a_n^2$. That they are all positive means they are all equal. They all sum to $m$, so they all must me $m/n$.

Then sum their reciprocals to get $\frac{n^2}{m}$ as the minimum.