Volume 9, Issue 3

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Proofs of Inequalities

Let be real numbers. Prove that

for all integer .

Define .

First, we use NMaximize to examine the inequality for .

From the form and statement of the inequality, a proof by induction [mathworld.wolfram.com/PrincipleofMathematicalInduction.html] looks like the way to proceed. Leon Aigret (aigret@myrealbox.com) presented the following inductive proof on the alt.math.recreational newsgroup, here restated using Mathematica.

For , the proof is trivial.

We need to show that for all . Consider the transformation for . Here is the result with .

In this result we recognize involving instead of .

In general we have

So, assuming that , we need to prove that

It is straightforward to show that this inequality holds for and so we have completed the proof.

Andrzej Kozlowski (www.mimuw.edu.pl/~akoz) proved the following stronger statement:

for every positive integer .

It is easy to see that this statement implies the original inequality using Cauchy's inequality [mathworld.wolfram.com/CauchysInequality.html],

Setting ,

and hence

Moreover, the proof is easier since the inductive step is now trivial.

So, assuming that , we need to prove that

This is clearly true for .

If , then so the proof is complete.

In addition, the inequality leads to some intriguing observations since it implies that the sums, considered as functions on the real line, are bounded and attain their maxima. So it is natural to consider the functions obtained by setting all the .

The limit as can be computed in closed form.

It is interesting to visualize the convergence of .