generalized AM-GM inequality

As I was reading through my linear analysis notes today, there was a small
passage where our lecturer Jim Wright provided a proof of a generalisation
of the inequality of arithmetic and geometric means:

Let $A, B \geq 0$ and $0 \leq \theta \leq 1$. Then $A^{\theta} B^{1-\theta} \leq \theta A + (1-\theta) B$.

I didn’t bother to read the proof as I thought I could probably come up with a
good one myself. Indeed, I am sufficiently happy with the proof I found that
I think it’s worth sharing. Here it is:

Proof:

If $AB=0$ the proof is trivial. So I shall proceed by assuming the contrary:

1. Let’s define $\beta := \frac{A}{B} \in \mathbb{R}_{+}$ and divide by $B$. We then obtain:$\beta^{\theta} \leq \theta \beta + (1-\theta)$
2. Now, let’s define the following differentiable functions:$\begin{cases} f(x) =\theta x + (1-\theta) \\ g(x) =x^{\theta} \\ \end{cases}$
3. The derivatives of these functions are given by:$\begin{cases} f'(x) =\theta \\ g'(x) =\theta x^{\theta-1} \\ \end{cases}$
4. Now, we can quickly reach the following conclusions:a) $f(x) \geq g(x)$ for $x \geq 1$
b) For $x \in [0,1]$ $\begin{cases} f(0) \geq g(0)=0 \\ f(1)=g(1)=1 \\ \end{cases}$

and both functions are monotonically increasing on $(0,1)$ so there can’t be $x \in (0,1)$ such that $f(x) = g(x)$ unless $\theta$ equals zero or one.

It follows that $f(x) \geq g(x)$ on $(0,1)$.

I must say that this was an easy problem but I liked the method I found for
solving it. Namely, reducing the number of variables and then replacing
variables with functions that can then be readily analysed. But, we can go a bit further
and show how Hölder’s inequality follows easily.

Hölder’s inequality:

For any two sequences $(a_i)_{i=1}^n,(b_i)_{i=1}^n \subset \mathbb{N}$, we have:

$\sum_{j=1}^{n} a_j b_j = 1 \leq (\sum_{j=1}^{n} a_j^p)^{\frac{1}{p}}(\sum_{j=1}^{n} b_j^q)^{\frac{1}{q}}$

for $1 \leq p \leq \infty$ where $\frac{1}{p} + \frac{1}{q}=1$.

Proof:

If we define $a=(a_i)_{i=1}^n, b= (b_i)_{i=1}^n$, and normalize these vectors we have:

$\begin{cases} \alpha = \frac{a}{||a||_p} \\ \beta = \frac{b}{||b||_q} \\ \end{cases}$

Now, we may apply the generalised AM-GM inequality to deduce:

$\sum_{j=1}^{n} \alpha_j \beta_j \leq \frac{1}{p}(\sum_{j=1}^{n} \alpha_j^p) +\frac{1}{q}(\sum_{j=1}^{n} \beta_j^q)= ||\alpha||_p||\beta||_q$

One thought on “generalized AM-GM inequality”

1. Interesting proof