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# 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
3 years, 3 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 $$a_1, a_2, \cdots$$ 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. · 3 years, 3 months ago

I will be happy to post my solutions if requested (as long as you try them first!) · 3 years, 3 months ago

Could you add these examples to the Applications of Cauchy Schwarz Inequality Wiki page? Staff · 2 years, 5 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 $$\frac{x+y+z}{3} \geq \sqrt[3]{xyz}$$, 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. · 3 years, 3 months ago

Could you explain the second one as in how did you use Cauchy Schwartz in it. · 2 years, 1 month ago

What are symmetric inequalities? · 2 years, 8 months ago

Hope you share many more such techniques!!thnx a lot for this. · 3 years, 3 months ago

Yup I will. In fact, I posted some on invariants and I also wrote some worked examples in my newest post. :) · 3 years, 3 months ago

@Anqi Li Can you add this to the Cauchy Schwarz Inequality Wiki page? Staff · 2 years, 5 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. · 3 years, 3 months ago