Lucas Willems

LUCAS WILLEMS

A 21 year-old student passionate about maths and programming

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Cauchy-Schwarz inequality for sums

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The Cauchy-Schwarz inequality for sums is the following :

$$\left(\sum_{k = 1}^{n} a_k b_k\right)^2 \leq \left(\sum_{k=1}^{n}a_k^2\right) \left(\sum_{k=1}^{n}b_k^2\right)$$

The purpose of this article is to prove this inequality.

Summary

Proof

The proofs consists of studying the following function :

$$f(x) = \sum_{k=1}^{n} (a_k x + b_k)^2$$

The first thing we can notice is that :

$$(a_k x + b_k)^2 \geq 0 \quad \Rightarrow \quad f(x) \geq 0 \quad \forall x \in \mathbb{R}$$

Then, let's try to develop \(f(x)\) :

$$\begin{align} f(x) &= \sum_{k=1}^{n} (a_k^2 x^2 + 2a_k b_k x + b_k^2) = \sum_{k=1}^{n} a_k^2 x^2 + \sum_{k=1}^{n} 2 a_k b_k x + \sum_{k=1}^{n} b_k^2 \\ &= \left(\sum_{k=1}^{n} a_k^2\right) x^2 + 2\left(\sum_{k=1}^{n} a_k b_k\right) x + \sum_{k=1}^{n} b_k^2 \end{align}$$

The 2nd thing we can notice is that \(f(x)\) is quadratic. Furthermore, as we have just seen that \(f(x)\) is also positive, \(\Delta \leq 0\). So :

$$\left(2\left(\sum_{k=1}^{n} a_k b_k\right)\right)^2 - 4 \left(\sum_{k=1}^{n} a_k^2\right) \left(\sum_{k=1}^{n} b_k^2\right) \leq 0 \quad \Rightarrow \quad 4\left(\sum_{k=1}^{n} a_k b_k\right)^2 \leq 4 \left(\sum_{k=1}^{n} a_k^2\right) \left(\sum_{k=1}^{n} b_k^2\right)$$

We just have to divide by 4 and we get the famous Cauchy-Schwarz inequality for sums !

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