How to prove that $1-e^{-\lambda}\leq \lambda ^{\alpha}$ for all $\lambda\geq 0$ and $\alpha\in (0,1]$.

Question

How to show that $1-e^{-\lambda}\leq \lambda ^{\alpha}$ for all $\lambda\geq 0$ and $\alpha\in (0,1]$.

My attempt: Let us define $f(\lambda)= \lambda^{\alpha}+e^{-\lambda}-1$ then $f'(\lambda)= \alpha\lambda^{\alpha-1}-e^{-\lambda}=\alpha \lambda^{\alpha}\frac{1}{\lambda}-\frac{1}{e^{\lambda}}$ and using that $\alpha<1$ and $-\frac{1}{e^{\lambda}}$ is negative then

$f'(\lambda)\leq \lambda^{\alpha}\frac{1}{\lambda}$ i believe that i have eproblems with $\lambda\in (0,1)$. For $\lambda>1$ then $f'(\lambda)\leq \lambda^{\alpha}$

The other way that i was tried is by series $1-e^{-\lambda}=\sum_{n=1}^{\infty}(-1)^{n+1}\frac{\lambda ^n}{n!}$ but i am stuck. How can continued please?

Thank you

Answer 1

For $0 \le \lambda \le 1$ is $$ 1 - e^{-\lambda} \le \lambda \le \lambda^\alpha \, , $$ for the first inequality see for example Simplest or nicest proof that $1+x \le e^x$.

And for $\lambda > 1$ is $$ 1 - e^{-\lambda} < 1 < \lambda^\alpha \, . $$

The following plot of $1-e^{-x}$ (blue), $\min(1, x)$ (purple), $x^{1/2}$ (red) and $x^{1/4}$ (green) shows the idea: