I've seen the following and I'm not sure whether it is true or not, and if yes, why it holds.
$(1-p)^x \geq 1-p x$ for $p\in (0,1)$ and $x>0$.
Do I need some additional Information to prove that?
We have $$(1-p)^x = \sum_{n=0}^{\infty} { x \choose n} \cdot (-p)^n=1-px + \sum_{n=2}^{\infty} { x \choose n} \cdot (-p)^n$$.
Now we would need to Show that $$\sum_{n=2}^{\infty} { x \choose n} \cdot (-p)^n \geq 0$$, then everything would be great. But I'm not sure whether this is possible.
Thank you very much for your help
It's false for $x=p=\frac{1}{2}$. However, it's true for $x\geq1$. Indeed, you can study $p\rightarrow(1-p)^x-1+px$, the derivative is positive.