How to prove $e^t-1 > (e-1) t^2$.

Question

I want to prove

$$e^t-1 \ge (e-1) t^2, \qquad t\in[0,1].$$

For $t\in[0,1]$, we have $t^n \le t^m$ whenever $m>n.$ I want to use this inequality in the proof of above statement but unable to proceed for terms beyond $t^2.$

Answer 1

The inequality is trivially an equality for $t=0$ or $t=1$. Define $$f(t):=\big(\exp(t)-1\big)-(\text{e}-1)\,t^2=\sum_{k=1}^\infty\,\frac{t^k}{k!}-\sum_{k=1}^\infty\,\frac{t^2}{k!}=(t-t^2)-\sum_{k=3}^\infty\,\frac{t^2(1-t^{k-2})}{k!}$$ for $t\in(0,1)$. That is, $$g(t):=\frac{f(t)}{t(1-t)}=1-\sum_{k=3}^\infty\,\frac{t\,\left(1+t+t^2+\ldots+t^{k-3}\right)}{k!}> 1-\sum_{k=3}^\infty\,\frac{(k-2)}{k!}$$ for every $t\in(0,1)$. Now, $$\sum_{k=3}^\infty\,\frac{(k-2)}{k!}=\sum_{k=3}^\infty\,\frac{1}{(k-1)!}-2\,\sum_{k=3}^\infty\,\frac{1}{k!}=\left(\text{e}-2\right)-2\left(\text{e}-\frac{5}{2}\right)=3-\text{e}\,.$$ That is, $$g(t)> 1-(3-\text{e})=\text{e}-2>0\,.$$ The claim immediately follows. In fact, this proof gives a stronger inequality: $$\exp(t)\geq 1+ (\text{e}-1)\,t^2+(\text{e}-2)\,t(1-t)=1+(\text{e}-2)\,t+t^2\text{ for every }t\geq 0\,.$$

Answer 2

Alternative (and straightforward) approach. Note that $f(t)=e^t-1 -(e-1) t^2$ is concave in $[0,1]$ because for $t\in[0,1]$, $$f''(t)=e^t-2(e-1)\leq e-2e+2=-e+2<0$$ and $f(0)=f(1)=0$. Hence, for $t\in[0,1]$, $$e^t-1 -(e-1) t^2=f(t)=f(0+t\cdot 1)\geq f(0)+t\cdot f(1)=0.$$

Answer 3

Hint:

By the transformation

$$x(1-x)=z,$$ or

$$x=\frac{1\pm\sqrt{1-4z}}2$$ we fold the axis and merge the two roots at $z=0$.

Then the two branches of the function

$$f(z):=e^{\frac{1\pm\sqrt{1-4z}}2}-1-(e-1)\left(\frac{1\pm\sqrt{1-4z}}2\right)^2$$

seems to have a Taylor development with positive-only coefficients.

But this needs to be proven.

Answer 4

Proof

Denote $$f(x)=x^t,~~~~~t \in [0,1].$$

Then by Lagrange's Mean Value Theorem, we have $$f(e)-f(1)=f'(\xi)(e-1),~~~\xi \in (1,e).$$

Namely, $$e^t-1=t\xi^{t-1}(e-1).$$

Thus, $$\xi^{t-1} \geq e^{t-1} \geq t\geq 0,$$which is desired.