# Strange proof of Young's inequality

1. Jan 16, 2012

### AxiomOfChoice

I'm trying to absorb a perplexing proof of Young's inequality I've found. Young's inequality states that if $A,B \geq 0$ and $0 \leq \theta \leq 1$, then $A^\theta B^{1-\theta} \leq \theta A + (1-\theta)B$.

The first step they take is the following: We can assume $B \neq 0$. (I get that.) But then they say we can just replace $A$ with $AB$, in which case it suffices to prove $A^\theta \leq \theta A - (1-\theta)$.

I understand why the replacement of A with AB in Young's inequality gets you to the new inequality in only A. But I don't see why you get to do that. Why is proving $A^\theta \leq \theta A - (1-\theta)$ logically equivalent to proving $A^\theta B^{1-\theta} \leq \theta A + (1-\theta)B$?

2. Jan 16, 2012

### AxiomOfChoice

Wow. Don't I feel dumb. Suppose you know that for $A \geq 0$ and $0 \leq \theta \leq 1$, $A^\theta \leq \theta A + (1-\theta)$. Then if $B > 0$, you know $A/B \geq 0$, so you have

$$\left( \frac{A}{B} \right)^\theta \leq \theta \frac{A}{B} + (1-\theta).$$

Rewriting this and multiplying both sides by $B$, you obtain

$$A^\theta B^{1-\theta} \leq \theta A + (1-\theta)B,$$

which is the desired result. So, I'm good to go. But I still think the book's way of explaining this bit of subterfuge ("replace A by AB") is poor.

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