Cauchy-Shwarz inequality is as follows:
Theorem. \[\left(\sum a_i^2\right) \left(\sum b_i^2\right) \ge \left(\sum a_i b_i \right)^2\] with equality iff all \(a_i\) are identical, and all \(b_i\) are also identical.
To see why this is true, first note that equality indeed holds if there are \(a,b\) so that all \(a_i = a\) and all \(b_i = b\). Then, note that perturbations from this equality case increase the left hand side more.
ok, but now here’s the question from the book:
If we are only a little bit away from the equality case, can we give some sort of statement of the form “all \(a_i\)’s and \(b_i\)’s are pretty close to each other”?
More formally this is the problem:
Proposition. Fix \(n\ge 2\). Let \(a_1,\ldots, a_n\) be positive numbers satisfying \[\sum a_i \sum \frac{1}{a_i} \le (n+1 / 2)^2.\] Then, \(\max a_i \le 4 \min a_i\).
Proof. Assume for sake of contradiction that the proposition is false. Fix \(a_i\) constituting a counter example. Let \(x=\min a_i\) and let \(\epsilon\) be such that \(\max a_i = (4+\epsilon)x\).
For notational convenience let \(a_{n-1}=x, a_{n} =(4+\epsilon)x\).
Note that Cauchy shwarz says \[\left(\sum a_i\right) \sum \frac{1}{a_i} \ge n^{2},\] so we are pretty darn close to the equality case. This is why we suspect it is reasonable to find a bound on the multiplicative size of the range of \(a_i\).
We have \[\left(\sum a_i\right) \left(\sum \frac{1}{a_i}\right) = \left((5+\epsilon)x+ \sum_{i=1}^{n-2} a_i\right) \left(\frac{1}{x} + \frac{1}{(4+\epsilon)x} + \sum_{i=1}^{n-2} \frac{1}{a_i}\right)\]
By Cauchy Shwarz: \[\left((5+\epsilon)x+ \sum_{i=1}^{n-2} a_i\right) \left(\frac{1}{x} + \frac{1}{(4+\epsilon)x} + \sum_{i=1}^{n-2} \frac{1}{a_i}\right) \ge \left(n-2 + \sqrt{(5+\epsilon)(1+\frac{1}{4+\epsilon})}\right)^{2}.\] Optimizing the quadratic, we find that for apporpriately chosen \(\epsilon\) the expression would be larger than \((n+ 1/2)^{2}\), a contradiction.