---

Source. Published p. 268, Proposition 2.3 (PDF).

Statement. Let $0<\delta\le1/10$ and $x,y\ge0$ satisfy $(1+y)^2\le1+\delta$ and $(1+x)(1-y)\le1+\delta$. Then

$$(1+y)(1-x)>1-\delta-2\delta^2.$$

Proof. The first assumption gives $y\le\delta/2<1$ and the second gives $x\le(\delta+y)/(1-y)$. The desired left side decreases with $x$, so it suffices to use that upper endpoint. Multiplying by $1-y$, the remaining inequality is

$$2\delta^2(1-y)>2(\delta+y)y.$$

The left side decreases and the right side increases as $y$ increases. At $y=\delta/2$ they are respectively $2\delta^2(1-\delta/2)$ and $3\delta^2/2$; the former is larger because $\delta<1/2$. This proves the result, including $x=0$ or $y=0$. $\square$