We all know the very famous Cauchy-Schwaz inequality (of the integral
form):
Let and be real functions continuous in . Then the following equality holds:
The purpose of this blog is to prove an enhanced version of this
inequality, which is stated as follows:
Let , , and . Prove that
Step 1: Substract from
From the vanilla C-S inequality, we know that if and are linear, i.e., , the equality holds. However in this
problem, we know that there does not exist such that because and has non-positive values in . This suggests us that, to reach
its minimum value as much as possible, we should manage to push to be as linear as as possible. To this end, we can
substract a constant , from , so that and can hold the same sign at any given
input .
So what should we choose for .
An intuitive choice is the average of , i.e., ,
which lies within the minimum value and maximum value .
A more important property of substracting a constant number from is, it does not change the value of
as
is given.
Then it follows that:
The last inequality applies the standard C-S inequality.
Now all we need to do is to prove that:
Step 2:
Discriminant of a quadratic equation
Rearranging the above inequality, we have:
It is clear that if we treat , and , we obtain the discriminant of
a quadratic equation:
We can construct such quadratic equation:
Then is equivalent to
fact that there exists at least one point on where . The last of our work is to
find such point.
Step 3: Find a point
The easiest way to find such point that the value of on that point is non-positive is by
trying some common values. For example, when , . When , we have:
It is a nice form but still not what we expect as the result is
positive. However, if we change the sign of , that is to turn to : , the new
will become:
This is exactly what we want: a value on that is non-positive. Now is:
and implies the
discriminant , which
closes our proof.