Proving that if $\forall \varepsilon > 0, a > 1 - \varepsilon^2$ then $\forall\varepsilon>0,a>1−2\varepsilon$, where $a \in \mathbb R$/.

Question

How to prove that, if $\forall \varepsilon > 0, a > 1 - \varepsilon^2$ then $\forall\varepsilon>0,a>1−2\varepsilon$, where $a \in \mathbb R$/.

I tried to manipulate the inequality, but I couldn't find a way of showing it. Conceptually, I know that this is right. Essentially, the first inequality is saying that $a$ is, at the very least, equal to 1. Thus, anything you subtract from 1 and you will give you something less than a. Right?

Answer 1

Fix $\varepsilon>0$, and put $\delta=\sqrt{2\varepsilon}$. Then $\delta>0$, so by hypothesis, $$a>1-\delta^2=1-2\varepsilon.$$ Since $\varepsilon>0$ was arbitrary, the result follows.

Answer 2

As already mentioned in the comments (@Ted Shifrin), both statements are equivalent to $a\geq 1$.

How to see that? Assume $a <1 \Rightarrow \exists c>0:\; a<c<1 $. Choose either

• $\varepsilon = \sqrt{1-c} \Rightarrow a \color{blue}{>} 1 - \varepsilon^2 = 1-(1-c) = c \color{blue}{>} a$ Contradiction! OR
• $\varepsilon = \frac{1-c}{2} \Rightarrow a \color{blue}{>} 1 - 2\varepsilon = 1-2\frac{1-c}{2} = c \color{blue}{>} a$ Contradiction!