Cauchy's inequality for olympiads

This is part of a series of articles based on my high school research paper from 2021 covering nonstandard approaches to solving olympiad problems. The problems from the Latvian Open Mathematics Olympiad mentioned in the article can be found in the LU NMS archive.

Problems where we have to either figure out the minimum or maximum value of an expression, or ones where inequalities are involved in some other way, are common in olympiads. Once helpful inequality at the Latvian Open Mathematics Olympiad, which the absolute majority of contests certainly do not even know, is Cauchy’s inequality.

Let’s get to know this inequality and see how it can be used to solve olympiad problems.

Cauchy’s inequality

Cauchy’s inequality is fairly simple, though, in my view, it’s easy to forget. In particular, it is easy to confuse which way the inequality is pointing, so that is worth memorising carefully. Here is the definition:

For any real numbers $r_1,\ldots,r_n$ and $s_1,\ldots,s_n$ we have \[(r_1^2+\ldots+r_n^2)(s_1^2+\ldots+s_n^2)\geq(r_1s_1+\ldots+r_ns_n)^2.\]

We will be able to replace $r_1,\ldots,r_n$ and $s_1,\ldots,s_n$ with basically any values or expressions.

Olympiad examples

Latvian Open Mathematics Olympiad 2016/2017, Form 10, Problem 2.
Prove that, for any positive numbers $a$ and $b\,$ we have $(\tfrac{3a}{b}+1)(\tfrac{3b}{a}+1)\geq16$.

First, a bit of motivation. We can see that each set of parentheses has two terms inside. To use Cauchy’s inequality, we would need \[r_1^2=\tfrac{3a}{b},\quad r_2^2=1,\quad s_1^2=\tfrac{3b}{a},\quad s_2^2=1.\] This means we could try substituting in \[r_1=\sqrt{\tfrac{3a}{b}},\quad r_2=1,\quad s_1=\sqrt{\tfrac{3b}{a}},\quad s_2=1.\] Since we know that $a$ and $b\,$ are positive, we can safely introduce square roots like this.

Substituting \[r_1=\sqrt{\tfrac{3a}{b}},\quad r_2=1,\quad s_1=\sqrt{\tfrac{3b}{a}},\quad s_2=1\] into Cauchy’s inequality gives \[(\tfrac{3a}{b}+1)(\tfrac{3b}{a}+1)\geq(\sqrt{\tfrac{3a}{b}}\cdot\sqrt{\tfrac{3b}{a}}+1\cdot1)^2.\] Since \[\sqrt{\tfrac{3a}{b}}\cdot\sqrt{\tfrac{3b}{a}}+1\cdot1=\sqrt{\tfrac{3\cancel a}{\cancel b}\cdot\tfrac{3\cancel b}{\cancel a}}+1=\sqrt{9}+1=4,\] the right side evaluates to $4^2=16$, hence we have shown the inequality.

Latvian Open Mathematics Olympiad 2016/2017, Form 11, Problem 2.
Given four positive numbers $a_1,a_2,a_3$ and $a_4$ such that $a_1a_3=a_2a_4=2017$, what is the smallest possible value of $(a_1+a_2)(a_3+a_4)$?

Substituting \[r_1=\sqrt{a_1},\quad r_2=\sqrt{a_2},\quad s_1=\sqrt{a_3},\quad s_2=\sqrt{a_4}\] into Cauchy’s inequality gives \begin{aligned} (a_1+a_2)(a_3+a_4)&\geq(\sqrt{a_1a_3}+\sqrt{a_2a_4})^2 \\ &=(\sqrt{2017}+\sqrt{2017})^2 \\ &=(2\cdot\sqrt{2017})^2 \\ &=4\cdot2017 \\ &=8068. \end{aligned} The value $8068$ is achieved, for instance, when $a_1=a_3=2017$ and $a_2=a_4=1$, therefore it is the smallest possible value.

Latvian Open Mathematics Olympiad 2016/2017, Form 12, Problem 2.
Prove that $\tfrac1a+\tfrac4b+\tfrac{16}c\geq\tfrac{49}{a+b+c}$ if $a,b\,$ and $c$ are positive numbers!

Here it helps to first notice that we are allowed to rewrite this inequality as \[(\tfrac1a+\tfrac4b+\tfrac{16}c)(a+b+c)\geq49.\] Then, this becomes reminiscent of the Form 10 problem from earlier.

Substituting \[r_1=\sqrt{\tfrac1a},\quad r_2=\sqrt{\tfrac4b},\quad r_3=\sqrt{\tfrac{16}c},\quad s_1=\sqrt a,\quad s_2=\sqrt b,\quad s_3=\sqrt c\] into Cauchy’s inequality gives \begin{aligned} (\tfrac1a+\tfrac4b+\tfrac{16}c)(a+b+c)&\geq(\sqrt{\tfrac1a}\cdot\sqrt a+\sqrt{\tfrac4b}\cdot\sqrt b+\sqrt{\tfrac{16}c}\cdot\sqrt c)^2 \\ &=(\sqrt1+\sqrt4+\sqrt{16})^2 \\ &=(1+2+4)^2 \\ &=7^2 \\ &=49. \end{aligned}Dividing both ends by $a+b+c$ returns the necessary inequality.

Final thoughts

From these problems, we can see that Cauchy’s inequality is helpful if we have a product of two sets of parentheses, combined with a weak inequality or a search for the minimum value.

Such problems have other approaches, too, but Cauchy’s inequality cracks open the problem instantly, hence it is definitely worth being aware of it.

This inequality also appears frequently in selection contests or international competitions, hence it is a good gateway to tougher algebra problems.