$$ e^x>2\cdot (e^x-\frac{x^{100}+100x^{99}+100\cdot99\cdot x^{98}+...+100!\cdot x+100!}{100!}) $$ proof that it's true for every x in [0,100]
I was trying to use maclaurin series of $e^x$ and after few steps got something like $$ 1+x+x^2/2!+x^3/3!+...+x^{100}/100!-x^{101}/101!-x^{102}/102!-....>0 $$ I have no idea what to do next:(
You just need to use the remainder in Taylor Series... in fact $$ e^x - \frac{x^{100}+ \cdots + 100!}{100!}= e^x - \sum_{n=0}^{100}\frac{x^n}{n!} = \frac{e^{\xi}}{101!}x^{101}, \quad \xi \in [0,x] $$
so you proof reduces to show that
$$ e^x > \frac{ 2 e^{\xi} x^{101}}{101!}. $$