$\frac 1 {1 + \epsilon} \le 1 - \frac \epsilon 2$ for $\epsilon \in (0, \frac 1 2)$

Question

How can we show that the following holds for $\epsilon \in (0, \frac 1 2)$? $$ \frac 1 {1 + \epsilon} \le 1 - \frac \epsilon 2 $$

I thought, maybe it would be more convenient to try to show somehow that $\frac 1 {1 + \epsilon} + \frac \epsilon 2 \le 1$. And maybe use the fact that $\frac {1 + \epsilon} {1 - \epsilon} > 1 + 2 \epsilon$.

Best regards

As a rule of our community, I request you to please show your efforts in solving it. Thanks. — Knight wants Loong back, Jul 17 '20 at 15:11

score 0 · Accepted Answer · answered Jul 17 '20 at 15:51

You can show this directly (and even for a wider interval for $\epsilon$) by rearranging the inequality:

$$\frac 1 {1 + \epsilon} \le 1 - \frac \epsilon 2 \Leftrightarrow 1\leq 1+\epsilon - \frac \epsilon 2(1+\epsilon) = 1 + \frac 12 \epsilon(1-\epsilon)$$

Since $\epsilon(1-\epsilon) \geq 0$ for $\epsilon \in [0,1]$, you are done.

$\frac 1 {1 + \epsilon} \le 1 - \frac \epsilon 2$ for $\epsilon \in (0, \frac 1 2)$

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