Show that for $\alpha, \beta > 0$ and $0 < \delta < 1$, that $\left| \alpha\beta -1 \right| \leq 3\delta$

Question

I am working on a pretty long proof, which is irrelevant here. But in one step, I need to prove the following:

Let $\alpha > 0$, $\beta > 0$ and $0 < \delta < 1$ be real values with $\left|\alpha - 1\right| \leq \delta$ and $\left|\beta - 1\right| \leq \delta$. Show that: $$\left|\alpha\beta -1 \right| \leq 3\delta$$

I'm sadly stuck and don't even know where to start here, which is holding me back in my proof. How can this be proven?

Answer 1

$$|ab-1| \leq |ab-a+a-1|\leq |a||b-1|+|a-1|$$ $$\leq (|a-1|+1)|b-1|+|a-1|$$ $$\leq \delta^2+\delta+\delta <3\delta$$

Answer 2

From $|\alpha - 1| \leq \delta$ and $|\beta - 1| \leq \delta$, we have $$ 1-\delta \leq \alpha \leq 1 + \delta $$ and $$ 1-\delta \leq \beta \leq 1 + \delta \text{,} $$ so $$ 1-2\delta + \delta^2 \leq \alpha \beta \leq 1+2\delta + \delta^2 \text{.} $$ Then $$ -2\delta + \delta^2 \leq \alpha \beta - 1\leq 2\delta + \delta^2 \text{.} $$ Since $0 < \delta < 1$, the triangle inequality gives $|-2 \delta + \delta^2| < 2 \delta + \delta^2 < 3 \delta$. So you have the slightly stronger $$ |\alpha \beta - 1| < 2\delta +\delta^2 < 3\delta \text{.} $$

Answer 3

(I'm using Latin letters rather than Greek because I'm lazy.)

$(a-1)(b-1) =ab-a-b+1 =ab-1-(a-1)-(b-1) $

so

$ab-1 =(a-1)+(b-1)+(a-1)(b-1) $.

Taking absolute values,

$|ab-1| \le 2d+d^2 \lt 3d $.