The AM - GM Inequality states that the arithmetic mean of a set of numbers $a_1,a_2...a_n$, will always be greater than or equal to the geometric mean of the same set.

More formally,

$\sum_{i=1}^n \frac{a_i}{n} \geq \sqrt[n]{\prod_{i=1}^n{a_i}}$

This has many applications, 1 of them specifically being to prove the Cauchy-Schwarz Inequality, which states that for a set $a_1,a_2...a_n$, and a set $b_1,b_2...b_n$, that the sum of the squares of each set multiplied together is greater than or equal to the square of the dot product of the two sets.

$\sum_{i=1}^n{a_i^2} \space \times \space \sum_{i=1}^n{b_i^2} \space\space\geq\space\sum_{i=1}^n{a_ib_i}$

Within my textbook, there is a proof for the CS inequality that applies the AM-GM inequality, and I am not completely sure what was done to achieve their result.

Below is the proof.

Let $A =\sqrt{\sum_{i=1}^n{a_i^2}} \space$ and $B =\sqrt{\sum_{i=1}^n{b_i^2}} \space$
Applying the AM-GM inequality, we have
$\sum_{i=1}^n{\frac{a_ib_i}{AB}} \leq \sum_{i=1}^n{\frac{1}{2}(\frac{a_i^2}{A^2}+\frac{b_i^2}{B^2})} = 1$
By cross multiplying...

The proof then continues into basic algebra to prove the inequality. My question, is where that crazy sigma equation came from.

  • $\begingroup$ Welcome to MSE. Please, be more careful choosing the tags. This is not a question of Measure Theory. $\endgroup$
    – jjagmath
    Apr 7, 2021 at 0:16
  • $\begingroup$ I apologize. I saw the word sigma when I searched Algebra and clicked. $\endgroup$
    – Smartsav10
    Apr 7, 2021 at 0:18
  • $\begingroup$ The LHS should be $\sum_{i=1}^n \frac{a_i b_i}{AB}$. Is that what is confusing you? $\endgroup$
    – jjagmath
    Apr 7, 2021 at 0:20
  • $\begingroup$ @jjagmath Thank you, but unfortunately no. I fixed it in the question. $\endgroup$
    – Smartsav10
    Apr 7, 2021 at 0:22

1 Answer 1


The mean in this case is between two terms: $a_i^2/A^2$ and $b_i^2/B^2$. If you write the AM-GM inequality for these terms for any given $i\in\{1,2,3...,n\}$, you get $\sqrt{ a_i^2/A^2 \times b_i^2/B^2 }\leq \frac{1}{2}(a_i^2/A^2 + b_i^2/B^2) $.

Summing both sides of this inequality across all $i$ preserves the inequality, and the resulting right hand side simplifies to one, giving the desired crazy sigma equation.

  • $\begingroup$ So are you saying that the terms are $\frac{a_1^2}{A^2}, \frac{a_2^2}{A^2}...\frac{a_n^2}{A^2}$. Also including the $b$ counterparts? $\endgroup$
    – Smartsav10
    Apr 7, 2021 at 0:28
  • $\begingroup$ For any particular $i\in \{1,2,3,...\}$, there are only two relevant terms involved in computing the means: $a^2_i/A^2$ and $b^2_i/B^2$. Write the AM-GM inequality for the mean of these two terms. Summing both sides of this inequality across all $i$ preserves the inequality. Does that make sense? $\endgroup$ Apr 7, 2021 at 0:31
  • $\begingroup$ Oh!!!!! I just realized that when I tried it out. It is multiple AMGM inequalities summed together. $\endgroup$
    – Smartsav10
    Apr 7, 2021 at 0:36
  • $\begingroup$ Essentially, yes-added more detail to my explanation $\endgroup$ Apr 7, 2021 at 0:37
  • $\begingroup$ Thank you so much! $\endgroup$
    – Smartsav10
    Apr 7, 2021 at 0:37

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