Reverse-deriving the rearrangement inequality
It always feels incredibly satisfying when a difficult task goes surprisingly smooth, right?
It might be due to the amount of mistakes I’ve made on my math exams, but I feel incredibly stressed when things work out smoothly. Almost as if something horrific is about to happen. A fakesolve. A miscalculation. Something worse than not knowing.
Today’s story takes a final look at the C-S inequality:
In honor of the C-S inequality, we’ll be rearranging things, using the C-S inequality to prove the rearrangement inequality, which is the opposite of what is usually done (using rearrangement to prove C-S).
The scary-looking omega is just a permutation of the numbers.
The Solution
It’s obvious that arranging similar numbers together in one term usually gives us larger values. If you don’t find it obvious, nobody will blame you; it’s simple to show, and achieved through a basic example.
Instead of dealing with symbolic math, you go back to numbers.
10 times 10 is 100, but 19 times 1 is just 19.
The numbers have the same sum, but distributing them more evenly achieves the larger product. Going back to the inequality, that corresponds to putting similar values closer to each other.
This is optimized when identical values are multiplied together, resulting in a sum of squares.
Now, you want to prove
Looks familiar, but not quite.
You have that aching feeling. A desire to write down the words “Evident by the C-S inequality”, but you can’t. What are you missing?
The C-S inequality has an overarching structure: squares and products of sums.
Right now, you only have individual squares and products of terms; these terms are obvious sums, so just try and square them.
The keen reader might notice a problem: the numbers are not restricted to nonnegative reals. However, if there are negative numbers, then it still holds, because the right hand side is always positive, and the left hand side would always be smaller with a negative term.
After squaring,
This is the C-S inequality. Equality holds if and only if
And we’re done. Interestingly, I did not make any false moves on this problem, so that’s all.
It’s that simple!
