# Understanding the proof of the Cauchy-Schwarz Inequality Using the AM-GM Inequality

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.

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

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.
• 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? Apr 7, 2021 at 0:28
• 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? Apr 7, 2021 at 0:31