Though underestimated and not-so-widely-used inequality; Rearrangement inequality is extremely useful in solving several Problems. Several inequalities can also be derived and proved easily from Rearrangement; such as AM-GM; Chebyshev’s inequality and so on…

Theorem

The sum

is maximal if Sequences and are similarly sorted; and minimal if sequences and are oppositely sorted.

**Main Idea**

If you see cyclic terms of the form then we can maximize or minimize it by rearrangement subject to given conditions. So we can write it into or of that structure. After forcing that structure, you can easily shift the terms that are in multiplied form and get a new structure; from which you can force identities or easier inequalities.

This idea might not be clear to the readers so I am giving us some examples of Rearrangement inequality.

Also, I will use frequently a notation :

and so on…

Theorems Derived from Rearrangement

We can prove the AM-GM inequality for numbers using rearrangement inequality. Here I present a short proof (from Problem solving strategies; Arthur Engel):

Let . Let

Define also:

We have the direct consequence

Sequences are oppositely sorted so that we easily have:

So we have derived AM-GM inequality. Next we will prove the Chebyshev’s inequality:

Theorem 2(Chebyshev)

(a)If sequences are similarly sorted then

And, (b)if sequences are oppositely sorted then we have:

Proof

We have;

We can prove (b) also accordingly using Rearrangement

Nessbitt’s inequality proved by rearrangement

For we have

We have already proved this using an elegant method by Titu’s lemma. We can also prove it using AM-GM inequality, and here I show my prof using Rearrangement:

Sequences and are similarly sorted(why?) so from Rearrangement inequality, we have:

So we are done.

Another problem

show that

We have, from easy rearrangement that

where the last one follows from AM-GM.

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Applications

Prove that where

Indeed we have :

Q.E.D.

Prove that

Of course;

Another argument

Let a,b and c be positive real numbers. Prove that

Taking logarithm of both sides to the base we have:

But, since sequences and are similarly sorted; and:

Hence from Rearrangement inequality, we obviously have:

Hence the result.

Let Prove that; if we always have:

Solution

What happens if we apply Cauchy like ?

Then we must prove that which is impossible. So direct Cauchy gives us a weak result.

What can we do? Rearrangement or AM-GM to the rescue. We rewrite the inequality as:

Now we just expand the LHS; so we are left to prove:

Which is perfectly true from Rearrangement or AM-GM inequality.

(AM-GM is applied directly here. Can you find a proof by direct rearrangement? )

Solution

This is a really old and well – known problem. My proof of this shows the full power of rearrangement.

There are several ways of proving this, but this is a new solution that I found, which, in my opinion, is the simplest and most elementary solution.

We rewrite the inequality as

Observe that sequences and are similarly sorted.

Hence from rearrangement;

So plugging this inequality into our required problem, we have

I finish my proof here .

(To be Updated)

For more problems, see my next entry on Chebyshev’s inequality