# Elementary Techniques used in the IMO (International Mathematical Olympiad) - Cauchy Schwarz Inequality

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! Note by Anqi Li
6 years, 1 month ago

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Hi 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 1995

Let $a, b, c \in \mathbb{R}^+$ 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 2009

For $n \geq 2$ 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.

- 6 years, 1 month ago

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

- 6 years, 1 month ago

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

Staff - 5 years, 3 months ago

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 $\frac{x+y+z}{3} \geq \sqrt{xyz}$, 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.

- 6 years, 1 month ago

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

- 4 years, 11 months ago

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

- 6 years, 1 month ago

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

- 6 years, 1 month ago

What are symmetric inequalities?

- 5 years, 6 months ago

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.

- 6 years, 1 month ago

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.

- 6 years, 1 month ago

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

Staff - 5 years, 3 months ago