If people asked you, what is the most elementary inequality you know? I bet your answer would be **AM-GM**. But in this series of posts I will try to show you the power that Cauchy Schwarz has over that of AM-GM. But as an introduction, let us first state and prove the theorem.

**Cauchy Schwarz Inequality**: Let $(a_1, a_2, \ldots , a_n)$ and $(b_1, b_2, \ldots, b_n)$ be two sequences of real numbers, then we have:

$(a_1^2 + a_2^2 + \ldots + a_n^2)(b_1^2 + b_2^2 + \ldots + b_n^2) \geq (a_1b_1 + a_2b_2 + \ldots + a_nb_n)^2$

In particular, equality holds iff there exists $k \in \mathbb{R}$ for which $a_i = k b_i$ for $i = {1, \ldots, n}$.

*Proof*: We will present $2$ proofs, one originating from analysis on the *equality case*, the other by wishful thinking on small cases of $n = 2,3$.

(i) Consider defining the following function $f$:

$f(x) = (a_1x - b_1)^2 + (a_2x - b_2)^2 + \ldots + (a_nx - b_n)^2$.

We will expand this to get:

$f(x) = (a_1^2 + a_2^2 + \ldots + a_n^2)x^2 - 2(a_1b_1 + a_2b_2 + \ldots a_nb_n)x + (b_1^2 + b_2^2 + \ldots + b_n^2)$

From our first way of representing $f(x)$, we can conclude that $f(j) ≥ 0 \leftrightarrow ∆_f \leq 0$ or

$(a_1^2 + a_2^2 + \ldots + a_n^2)(b_1^2 + b_2^2 + \ldots + b_n^2) \geq (a_1b_1 + a_2b_2 + \ldots + a_nb_n)^2$

Equality holds if the equation $f(x) = 0$ has one root.■

(ii) Just remark that:

$(a_1^2 + a_2^2 + \ldots + a_n^2)(b_1^2 + b_2^2 + \ldots + b_n^2) - (a_1b_1 + a_2b_2 + \ldots + a_nb_n)^2 = \sum_{i,j=1}^n (a_ib_j + a_jb_i)^2 ≥ 0$■

Let us note the following positive things regarding Cauchy-Schwarz:

it is effective in proving symmetric inequalities

Try to form squares

Helps to clear up square roots

Look forward to the next few posts to see applications of this extremely elegant inequality!

No vote yet

1 vote

Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in`\(`

...`\)`

or`\[`

...`\]`

to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewestHi everyone, I'm part of #TorqueGroup too. My assignment was to post olympiad level problems for the community that I found interesting. Since Anqi is posting these great topics about math things useful in such math, it seems like we're a perfect fit. So, here are some examples:

Show that for real numbers $a_1, a_2, \cdots$ and $b_1, b_2, \cdots$

$\displaystyle \sum_{i = 1}^{k} \frac {a_i^2}{b_i} \geq \frac{(\sum_{i = 1}^k a_i)^2}{\sum_{i = 1}^k b_i}$

IMO 1995Let $a, b, c \in \mathbb{R}^+$ such that $abc = 1$. Prove that

$\frac{1}{a^3(b+c)} + \frac{1}{b^3(a+c)} + \frac{1}{c^3(a+b)} \geq \frac{3}{2}$.

..

Hint: It may help to make a symmetric substitution for $a = \alpha, b = \beta, c = \gamma$ such that the condition $\alpha \beta \gamma = 1$ still holds.

USAMO 2009For $n \geq 2$ let $a_1, a_2, \cdots, a_n$ be positive reals such that

$(a_1 + a_2 + \cdots + a_n)(\frac{1}{a_1} + \frac{1}{a_2} + \cdots + \frac{1}{a_n}) \leq (n + \frac{1}{2})^2$

Prove that $max (a_1, a_2, \cdots, a_n) \leq 4 min (a_1, a_2, \cdots, a_n)$.

..

Hint: The LHS should scream cauchy. Choose which $a_i$ multiplies which $b_k$ to isolate a desired max and min, and then do some algebra.

Log in to reply

I will be happy to post my solutions if requested (as long as you try them first!)

Log in to reply

Could you add these examples to the Applications of Cauchy Schwarz Inequality Wiki page?

Log in to reply

For the second one, following your hint, let $a = \frac{1}{x}, b = \frac{1}{y}, c = \frac{1}{z}$. The expression is then $\sum_{cyc} \frac{x^3yz}{y+z}$. Since $xyz = 1$, this is $\sum_{cyc} \frac{x^2}{y+z}$. Using Cauchy, this is greater than $\frac{(x+y+z)^2}{2(x+y+z)}$. So now it is left to prove that $x+y+z \geq 3$. This is simple using AM-GM, so $\frac{x+y+z}{3} \geq \sqrt[3]{xyz}$, and we are done. $\blacksquare$

For the second one, WLOG assume $a_1$ is the maximum and $a_n$ is the minimum. Using Cauchy,

$(a_1 + a_2 + \cdots + a_n)(\frac{1}{a_1} + \frac{1}{a_2} + \cdots + \frac{1}{a_n}) \geq (\sqrt{\frac{a_1}{a_n}} + \sqrt{\frac{a_n}{a_1}} + n - 2)$

$(n + \frac{1}{2})^2 \geq (\sqrt{\frac{a_1}{a_n}} + \sqrt{\frac{a_n}{a_1}} + n - 2)^2$

Taking square roots of both sides, we get

$n + \frac{1}{2} \geq \sqrt{\frac{a_1}{a_n}} + \sqrt{\frac{a_n}{a_1}} + n - 2$

$\frac{5}{2} \geq \sqrt{\frac{a_1}{a_n}} + \sqrt{\frac{a_n}{a_1}}$

Now, squaring, we get

$\frac{25}{4} \geq \frac{a_1}{a_n} + 2 + \frac{a_n}{a_1}$

Multiplying both sides by $a_1 a_n$ we get

$a_1^2 - \frac{17}{4} a_1 a_n + a_n^2 \leq 0$

Factoring...

$(a_1 - 4a_n)(a_1 - \frac{1}{4}a_n) \leq 0$

Since $a_1$ and $a_n$ are positive reals, $(a_1 - 4a_n)$ must be nonpositive. Thus,

$(a_1 - 4a_n \leq 0 \rightarrow a_1 \leq 4a_n$, and we are done.

Log in to reply

Could you explain the second one as in how did you use Cauchy Schwartz in it.

Log in to reply

Hope you share many more such techniques!!thnx a lot for this.

Log in to reply

Yup I will. In fact, I posted some on invariants and I also wrote some worked examples in my newest post. :)

Log in to reply

What are symmetric inequalities?

Log in to reply

One can easily prove Cauchy-Schwarz using vectors too. It is similar to the proof above, but is much cleaner. The interesting this is that almost every other inequality can be derived from this inequality.

Log in to reply

Cauchy-Schwartz is indeed a remarkable inequality. But there are many other inequalities that are considerably more subtle and/or general. The first of them is probably the Holder's inequality which is a direct generalization of the Cauchy-Schwartz.

Log in to reply

@Anqi Li Can you add this to the Cauchy Schwarz Inequality Wiki page?

Log in to reply