# Lower bound for sum of reciprocals of positive real numbers

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).

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

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 $$n^2=n\left(a_1\frac{1}{a_1}+a_2\frac{1}{a_2}..+a_n\frac{1}{a_n}\right)\le \left(\sum_{i=1}^{n}a_i\right)\left(\sum_{i=1}^{n}\frac{1}{a_i}\right)$$

• @newman_ash were you able to follow the chebyshevs proof? Nov 22, 2020 at 10:33
• Yes I was able to follow it Nov 22, 2020 at 10:41
• But to me it is not obvious how to get rid of $\sum a_i^2$ and $\sum 1/a_i^2$ in the Cauchy-Schwarz inequality Nov 22, 2020 at 11:05
• @newman_ash i edited Nov 22, 2020 at 11:49

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

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.