Weekly Problem 2: Harmonic, Geometric And Arithmetic Mean Inequality

Theorem

This week the problem is one of the most beautiful proof that I have ever seen. There many proofs about this theorem, I will write here my favorite ones (in other words, I will write what I know).

Let \(x_1, x_2, x_3, ... , x_n \in \mathbb{R}^{+}\). Then, HM (Harmonic mean) \(\leq\) GM (Geometric mean) \(\leq\) AM (arithmetic mean), in mathematical notation:

\[\frac{n}{\frac{1}{x_1} + \frac{1}{x_2} + ... + \frac{1}{x_n}} \leq \sqrt[n]{x_1x_2...x_n} \leq \frac{x_1+x_2+...+x_n}{n}\]

, with equality in both cases if and only if all \(x_i\)’s are equal. Before we prove this, try to prove yourself.

Proof 1

If you not familiar with mathematical induction see [1]. The proof 1 is showed by non-standard induction. Let \(P(n)\) be the statement of the second inequality, in other words, written in the form as well as,

\[x_1 x_2...x_n \leq \left(\frac{x_1+x_2+...+x_n}{n}\right)^{n}\]

For \(n=2\), we have \(x_1 x_2 \leq \left(\frac{x_1+x_2}{2}\right)^{2} \iff (x_1 + x_2)^2 \geq 0\), that is true. Now we have to show two things:

• (1) \(P(n) \implies P(n-1)\)
• (2) \(P(n)\) and \(P(2) \implies P(2n)\)

which will clearly show us the full result.

To prove (1), let \(A = \sum_{k=1}^{n-1}\frac{x_k}{n-1}\), then

\[\left(\prod_{k=1}^{n-1}x_k\right)A \overset{\mathrm{P(n)}}{\leq} \left(\frac{\sum_{k=1}^{n-1}x_k + A}{n}\right)^2\] \[\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=\left(\frac{(n-1)A+A}{n}\right)^n = A^{n}\]

and hence,

\[\prod_{k=1}^{n-1}x_k \leq A^{n-1} = \left(\frac{\sum_{k=1}^{n-1}x_k}{n-1}\right)^{n-1}\]

To prove (2), we see

\[\prod_{k=1}^{2n}x_k = \left(\prod_{k=1}^{n}x_k\right)\left(\prod_{k=n+1}^{2n}x_k\right)\] \[\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\overset{\mathrm{P(n)}}{\leq}\left(\sum_{k=1}^{n}\frac{x_k}{n}\right)^n\left(\sum_{k=n+1}^{2n}\frac{x_k}{n}\right)^n\] \[\;\;\;\overset{\mathrm{P(2)}}{\leq}\left(\frac{\sum_{k=1}^{2n}\frac{x_k}{n}}{2}\right)^{2n}\] \[\;\;\; = \left(\frac{\sum_{k=1}^{2n}x_k}{2n}\right)^{2n}\]

The condition for equality is derived just as easily.

Now, let’s prove the inequality \(\frac{n}{\frac{1}{x_1} + \frac{1}{x_2} + ... + \frac{1}{x_n}} \leq \sqrt[n]{x_1x_2...x_n}\). Actually it is a direct consequence of \(x_1 x_2...x_n \leq \left(\frac{x_1+x_2+...+x_n}{n}\right)^{n}\). Now, we will define a new set that is related to \(x_1, x_2, x_3, ... , x_n\). \(y_j = \frac{1}{x_j}\) for \(j \in 1,2,..,n\). So, our GM (geometric mean) - AM (Arithmetic Mean) inequality becomes:

\[\sqrt[n]{\frac{1}{y_1 y_2...y_n}} \leq \frac{\frac{1}{y_1}+\frac{1}{y_2}+...+\frac{1}{y_n}}{n}\]

Taking the reciprocal of both sides yields,

\[\frac{n}{\frac{1}{y_1}+\frac{1}{y_2}+...+\frac{1}{y_n}} \leq \sqrt[n]{y_1 y_2...y_n}\]

as we desired.

\[Q.E.D\]

Proof 2

This proof uses Bernoulli’s inequality, that is:

\[(1+t)^{n+1} \geq 1+(n+1)t\;\;\;\;\;\;\;\;\; [\forall t \geq -1,\;\;\; t \in \mathbb{R}]\]

Suppose, \(x_1, x_2, x_3, ... , x_{n+1} > 0\) and let,

\[t = \frac{\frac{x_1+x_2+...+x_{n+1}}{n+1}}{\frac{x_1+x_2+...+x_n}{n}} - 1\]

By Bernoulli,

\[t = \left(\frac{\frac{x_1+x_2+...+x_{n+1}}{n+1}}{\frac{x_1+x_2+...+x_n}{n}}\right)^{n+1} \geq 1+(n+1)\left( \frac{\frac{x_1+x_2+...+x_{n+1}}{n+1}}{\frac{x_1+x_2+...+x_n}{n}}-1\right)\] \[=\;\;\;\;\; 1+n\frac{x_1+x_2+...+x_{n+1}}{x_1+x_2+...+x_n} - (n+1)\] \[=\;\;\;\;\; \frac{n x_{n+1}}{x_1+x_2+...+x_n}\]

which translates into:

\[\left(\frac{x_1+x_2+...+x_{n+1}}{n+1}\right)^{n+1} \geq x_{n+1}\left(\frac{x_1+x_2+...+x_n}{n}\right)^n\]

and the arithmetic-geometric mean inequality left to the reader (follows by induction).

\[Q.E.D\]