Strange proof of Young's inequality

[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?

 

[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

<br /> \left( \frac{A}{B} \right)^\theta \leq \theta \frac{A}{B} + (1-\theta).<br />

Rewriting this and multiplying both sides by B, you obtain

<br /> A^\theta B^{1-\theta} \leq \theta A + (1-\theta)B,<br />

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.