Cauchy Schwarz Inequality


The $Cauchy\ Schwarz\ Inequality$ states that for any given sequence $a_1,\ a_2\ ...,\ a_n$ and $b_1,\ b_2\ ...,\ b_n$, the following inequality holds:

$\sum\limits_{i=1}^{n}a_i^2\sum\limits_{i=1}^{n}b_i^2 \ge \left(\sum\limits_{i=1}^{n}a_ib_i\right)^2$

and in expanded form the inequality looks like:

$(a_1^2 + a_2^2 + ... + a_n^2) (b_1^2 + b_2^2 + ... + b_n^2) \ge (a_1b_1 + a_2b_2 + ... + a_nb_n)^2$

There are quite a few proofs of this theorem, and here we will see a polynomial based proof.

Let polynomial $p(x)$ be defined as:
$p(x) = (a_1x - b_1)^2 + (a_2x - b_2)^2 + ... + (a_nx - b_n)^2 = \sum\limits_{i = 1}^{n}(a_ix - b_i)^2$

Since $p(x)$ is the sum of squares of real numbers, $p(x)$ can either be positive or zero.
Let us see under which condition $p(x)$ will be $0$.
For $p(x)$ to be zero each term $(a_ix - b_i)$ has to zero. Therefore, we can say that for some value of $x$,

$\dfrac{b_i}{a_i} = x$
Therefore, the ratio of $a_i$ and $b_i$ must be constant for each $i$ for $p(x)$ to be $0$.

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Now let us consider the $p(x) \gt 0$ case.

$p(x) = \sum\limits_{i = 1}^{n}(a_ix - b_i)^2$

$= \sum\limits_{i = 1}^{n}(a_i^2x^2 - 2a_ib_ix + b_i^2)$

$= \sum\limits_{i = 1}^{n}a_i^2x^2 - \sum\limits_{i = 1}^{n}2a_ib_ix + \sum\limits_{i = 1}^{n}b_i^2$

$= x^2\sum\limits_{i = 1}^{n}a_i^2 - x\sum\limits_{i = 1}^{n}2a_ib_i + \sum\limits_{i = 1}^{n}b_i^2$

The above is a quadratic relationship in $x$. Since it is always positive or zero, it lies wholly above the $x$ axis touching $x$ axis at a single point, which gives us the single root for $p(x) = 0$ case as we saw before.
Therefore, the discriminant must be zero (for the one root case) or imaginary.

that is $\left(\sum\limits_{i = 1}^{n}2a_ib_i\right)^2 - 4\left(\sum\limits_{i = 1}^{n}a_i^2\right)\left(\sum\limits_{i = 1}^{n}b_i^2\right) \le 0$

$\Rightarrow 4\left(\sum\limits_{i = 1}^{n}a_ib_i\right)^2 - 4\left(\sum\limits_{i = 1}^{n}a_i^2\right)\left(\sum\limits_{i = 1}^{n}b_i^2\right) \le 0$

$\Rightarrow \left(\sum\limits_{i = 1}^{n}a_ib_i\right)^2 - \left(\sum\limits_{i = 1}^{n}a_i^2\right)\left(\sum\limits_{i = 1}^{n}b_i^2\right) \le 0$

$\Rightarrow \left(\sum\limits_{i = 1}^{n}a_ib_i\right)^2 \le \left(\sum\limits_{i = 1}^{n}a_i^2\right)\left(\sum\limits_{i = 1}^{n}b_i^2\right)$

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$\Rightarrow \left(\sum\limits_{i = 1}^{n}a_i^2\right)\left(\sum\limits_{i = 1}^{n}b_i^2\right) \ge \left(\sum\limits_{i = 1}^{n}a_ib_i\right)^2$

Now try the following problem:

--------- Reference to question: a7cdec27-ccac-455f-9fcd-cc705d0a39ad ---------

We can further generalise the above problem in the following way:

If $a^2 + b^2 + c^2 = 1$
The maximum value of $pa + qb + ra$ is $\sqrt{p^2 + q^2 + r^2}$