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 (a1,a2,,an) (a_1, a_2, \ldots , a_n) and (b1,b2,,bn) (b_1, b_2, \ldots, b_n) be two sequences of real numbers, then we have:

(a12+a22++an2)(b12+b22++bn2)(a1b1+a2b2++anbn)2 (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 kR k \in \mathbb{R} for which ai=kbi a_i = k b_i for i=1,,n i = {1, \ldots, n} .

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

(i) Consider defining the following function f f :

f(x)=(a1xb1)2+(a2xb2)2++(anxbn)2 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)=(a12+a22++an2)x22(a1b1+a2b2+anbn)x+(b12+b22++bn2) 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) f(x) , we can conclude that f(j)0f0 f(j) ≥ 0 \leftrightarrow ∆_f \leq 0 or

(a12+a22++an2)(b12+b22++bn2)(a1b1+a2b2++anbn)2 (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 f(x) = 0 has one root.■

(ii) Just remark that:

(a12+a22++an2)(b12+b22++bn2)(a1b1+a2b2++anbn)2=i,j=1n(aibj+ajbi)20 (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
5 years, 10 months 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 a1,a2,a_1, a_2, \cdots and b1,b2,b_1, b_2, \cdots

i=1kai2bi(i=1kai)2i=1kbi\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,cR+a, b, c \in \mathbb{R}^+ such that abc=1abc = 1. Prove that

1a3(b+c)+1b3(a+c)+1c3(a+b)32\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=α,b=β,c=γa = \alpha, b = \beta, c = \gamma such that the condition αβγ=1\alpha \beta \gamma = 1 still holds.

USAMO 2009

For n2n \geq 2 let a1,a2,,ana_1, a_2, \cdots, a_n be positive reals such that

(a1+a2++an)(1a1+1a2++1an)(n+12)2(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(a1,a2,,an)4min(a1,a2,,an)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 aia_i multiplies which bkb_k to isolate a desired max and min, and then do some algebra.

Michael Tong - 5 years, 10 months ago

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

Michael Tong - 5 years, 10 months ago

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

Calvin Lin Staff - 5 years ago

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For the second one, following your hint, let a=1x,b=1y,c=1za = \frac{1}{x}, b = \frac{1}{y}, c = \frac{1}{z}. The expression is then cycx3yzy+z\sum_{cyc} \frac{x^3yz}{y+z}. Since xyz=1xyz = 1, this is cycx2y+z\sum_{cyc} \frac{x^2}{y+z}. Using Cauchy, this is greater than (x+y+z)22(x+y+z)\frac{(x+y+z)^2}{2(x+y+z)}. So now it is left to prove that x+y+z3x+y+z \geq 3. This is simple using AM-GM, so x+y+z3xyz3\frac{x+y+z}{3} \geq \sqrt[3]{xyz}, and we are done. \blacksquare

For the second one, WLOG assume a1a_1 is the maximum and ana_n is the minimum. Using Cauchy,

(a1+a2++an)(1a1+1a2++1an)(a1an+ana1+n2)(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+12)2(a1an+ana1+n2)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+12a1an+ana1+n2n + \frac{1}{2} \geq \sqrt{\frac{a_1}{a_n}} + \sqrt{\frac{a_n}{a_1}} + n - 2

52a1an+ana1\frac{5}{2} \geq \sqrt{\frac{a_1}{a_n}} + \sqrt{\frac{a_n}{a_1}}

Now, squaring, we get

254a1an+2+ana1\frac{25}{4} \geq \frac{a_1}{a_n} + 2 + \frac{a_n}{a_1}

Multiplying both sides by a1ana_1 a_n we get

a12174a1an+an20a_1^2 - \frac{17}{4} a_1 a_n + a_n^2 \leq 0

Factoring...

(a14an)(a114an)0(a_1 - 4a_n)(a_1 - \frac{1}{4}a_n) \leq 0

Since a1a_1 and ana_n are positive reals, (a14an)(a_1 - 4a_n) must be nonpositive. Thus,

(a14an0a14an(a_1 - 4a_n \leq 0 \rightarrow a_1 \leq 4a_n, and we are done.

Mike Kong - 5 years, 10 months ago

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

Aayush Patni - 4 years, 7 months ago

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

Kishan k - 5 years, 10 months ago

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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. :)

Anqi Li - 5 years, 10 months ago

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What are symmetric inequalities?

Isabelle Quaye - 5 years, 2 months ago

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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.

Bob Krueger - 5 years, 10 months ago

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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.

Alexander Borisov - 5 years, 10 months ago

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

Calvin Lin Staff - 5 years ago

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