Given $0\leq \beta\leq \tau\leq 1$, prove $e^\beta - 1\leq (1+\tau)\beta$ and $1-e^{-\beta}\geq (1-\frac{\tau}{2})\beta$

Question

How to prove $$e^\beta -1 \leq (1+\tau)\beta$$ and $$1-e^{-\beta}\geq (1-\frac{\tau}{2})\beta$$ , provided that $0\leq \beta\leq \tau\leq 1$.
Many thanks!

Answer 1

As suggested by Kavi Rama Murthy in the comments, for the first one, we have that

$$e^x\le 1+x+ x^2$$

and then

$$e^\beta \leq 1+\beta+\beta^2 \implies e^\beta-1 \leq (1+\beta)\beta\le (1+\tau)\beta$$

For the second one, we have that

$$e^{-x}\le 1-x+\frac12 x^2$$

and then

$$e^{-\beta} \leq 1-\beta+\frac12\beta^2 \implies 1-e^\beta \geq \left(1-\frac12\beta\right)\beta \ge \left(1-\frac12\tau \right)\beta$$

To prove $e^x\le 1+x+ x^2$ let consider $f(x)=e^x-1-x-x^2\le0,\, x\in[0,1]$ then

• $f(0)=0$
• $g(x)=f'(x)=e^x-1-2x \le 0$

indeed

• $g(0)=0$
• $g(1)=-2$
• $g'(x)=e^x-2=0 \implies x=\ln 2$ which is a minimum

therefore the inequality holds.

To prove $e^{-x}\le 1-x+\frac12 x^2$ let consider $f(x)=e^{-x}- 1+x-\frac12 x^2 \le0,\, x\in[0,1]$ then

• $f(0)=0$
• $g(x)=f'(x)=-e^{-x}+1-x \le 0$

indeed

• $g(0)=0$
• $g(1)=-\frac1e$
• $g'(x)=e^{-x}-1\le 0 $

therefore the inequality holds.

Answer 2

Hint: Define $$f(\beta)=\ln(1+(1+\tau)\beta)-\beta$$ and use calculus