The following was an exercise in some MIT course notes on p-adic numbers and Hensel's lemma.
Let $f\in \Bbb Z_p[X]$. Suppose that $b$ is a simple root of $f$. Prove that for any $a\in \Bbb Z_p$: if $|a-b|_p<|f'(b)|_p$ then $|f(a)|_p<|f'(a)|_p^2$. Conclude that if no $a\in \Bbb Z_p$ satisfies $|f(a)|_p<|f'(a)|_p^2$, then $f$ does not have a simple root in $\Bbb Z_p$.
I am having trouble with the first part, which should be easy.
We have $$f(x) = f(a) + f'(a)(x-a) + g(x)(x-a)^2$$ for some polynomial $g$.
Plugging in $x=b$ gives $f(a)=f'(a)(a-b)-g(b)(a-b)^2$ so taking valuations gives $|f(a)|_p=|f'(a)(a-b)-g(b)(a-b)^2|_p$.
Now, $$ |f'(a)(a-b)|=|f'(a)|\cdot |a-b|<|f'(a)|\cdot |f'(b)| $$
How should I continue ? I guess I should use somewhere the ultrametric inequality, but I don't see where.