Proof of Cauchy–Schwarz inequality

Question

I was reading about the Cauchy–Schwarz inequality from Courant, Hilbert - Methods Of Mathematical Physics Vol 1 and I can not understand what they mean when they said the line that has been highlighted with red in the picture given below

I can not understand why a and b has to be proportional and why is this so crucial for the roots to be imaginary and why we want the roots to be imaginary in the first place.

Answer 1

Hint: The quadratic equation $ax^2+bx+c=0$ has real roots if and only if $b^2-4ac\ge 0$.

Added: The quadratic polynomial $\sum_1^n (a_ix+b_i)^2$ is a sum of squares. Thus this polynomial is always $\ge 0$.

Recall that a quadratic $ax^2+bx+c$, where $a\gt 0$, is always $\ge 0$ if and only if the discriminant $b^2-4ac$ is $\le 0$. Compute the discriminant of the messy quadratic. The inequality $b^2-4ac\le 0$ turns out to be precisely the C-S Inequality (well, we have to divide by $4$).

As to when we have equality, the quadratic has a real root $k$ if and only if $a_ik+b_i=0$ for all $i$. This is the case iff $b_i=-ka_i$, meaning that the $a_i$ and $b_i$ are proportional.

By the way, things are I think marginally prettier if we look at the polynomial $\sum_1^n (a_ix-b_i)^2$.

Answer 2

The expression $\displaystyle\sum_{i=1}^n (a_i x + b_i)^2$ cannot be negative since it's a sum of squares of real numbers, and cannot be zero unless every term is zero, in which case, since $a_i x=b_i$, the vectors $\vec{a}$ and $\vec{b}$ are proportional, and $x$ is the constant of proportionality.

Therefore the quadratic equation can have a real solution for $x$ only if the two vectors are proportional, in which case it has only one solution. If a quadratic equation with real coefficients has no real solutions or only one, then its discriminant is non-positive. In this case the discriminant is $$ \left(2\sum_{i=1}^n a_i b_i\right)^2 - 4\left(\sum_{i=1}^n a_i \right)\left(\sum_{i=1}^n b_i\right). $$ That expression must therefore be $\le0$, and $=0$ only if the two vectors are proportional. From that, the Cauchy-Scharz inequality follows.

Answer 3

We have the equation $$ \sum_{i=1}^n(a_ix+b_i)^2 =x^2\sum_{i=1}^na_i^2+2x\sum_{i=1}^na_ib_i+\sum_{i=1}^nb_i^2\tag{1} $$ If there were two distinct real roots of the right hand side of $(1)$, then the minimum of the quadratic would be at their average and would be less than $0$; this because the positive coefficient of $x^2$ yields strict convexity. However, if there were an $x$ so that the right hand side were less than $0$, the left hand side would yield a sum of squares that was negative.

$\Rightarrow$ the roots cannot be real and distinct

The only possibility for a real root would be in the case where the left hand side of $(1)$ were $0$. for that to occur, each term in the sum would need to be $0$. That is, for all $0\le i\le n$, $$ b_i=-a_ix\tag{2} $$ $\Rightarrow$ $a$ and $b$ must be proportional.

If the roots of the right hand side of $(1)$ are equal or non-real, we have, from the quadratic formula, that $$ \left(2\sum_{i=1}^na_ib_i\right)^2-\ 4\left(\sum_{i=1}^na_i^2\right)\left(\sum_{i=1}^nb_i^2\right)\le0\tag{3} $$ which upon rearrangement is Cauchy-Schwarz.

$\Rightarrow$ we want the roots of the right hand side of $(1)$ to be equal or non-real.