Could you please give me some hint on how to show (if possible) that the following inequality holds?
$$\bigg|(-1)^m\frac{t^m x^m}{m!}+\sum_{j=m+1}^{\infty}\frac{(-1)^jx^jt^j}{j!}\bigg|\leq \bigg|(-1)^m\frac{t^m x^m}{m!}\bigg|$$
(where $x>0$)
notice that the left side is nothing more than
$$\bigg|e^{-tx}-\sum_{k=0}^{m-1}\frac{(-1)^jx^jt^j}{j!}\bigg|$$
Any hint or advice will be really appreciated.
On
There exist useful estimates for the remainder in Taylor's expansion and that's precisely what you need. Here's your hint: https://en.wikipedia.org/wiki/Taylor%27s_theorem#Estimates_for_the_remainder
For $y\geqslant 0$ we have $1\geqslant e^{-y}=\sum_{k=0}^{\infty}(-1)^k y^k/k!$; integrating this $m$ times, we get $$\frac{y^m}{m!}\geqslant\sum_{k=0}^{\infty}\frac{(-1)^k y^{k+m}}{(k+m)!}\underset{k+m=j}{=}(-1)^m\sum_{j=m}^{\infty}(-1)^j\frac{y^j}{j!}.$$