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.
Best Answer
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.