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!

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## Comments

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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.

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

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

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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.

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

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What are symmetric inequalities?

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

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

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

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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.

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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.

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