My problem is the following:
Prove that $\left | e^{x}+e^{-x}-2-x^{2} \right |\leq \frac{1}{6}x^{4}$ when $\left | x \right|\leq1$.
Using Maclaurin's expansion formula, I get that
$$e^{x}=1+x+\frac{x^{2}}{2}+\frac{x^{3}}{6}+x^{4}\frac{e^{\Theta_1x}}{24},0\leq\Theta_1\leq1$$ $$e^{-x}=1-x+\frac{x^{2}}{2}-\frac{x^{3}}{6}+x^{4}\frac{e^{-\Theta_2x}}{24},0\leq\Theta_2\leq1$$
I can now simplify the inequality to: $$\left | x^{4}\frac{e^{\Theta_1x}}{24}+x^{4}\frac{e^{-\Theta_2x}}{24} \right |=\frac{x^{4}}{24}\left | (e^{\Theta_1x}+e^{-\Theta_2x}) \right |\leq \frac{1}{6}x^{4}\Leftrightarrow\frac{x^{4}}{4}\left | (e^{\Theta_1x}+e^{-\Theta_2x}) \right |\leq x^{4}$$
Aaaand here I'm pretty much stuck. What am I supposed to do here? I guess I have to find something in between my LHS and my RHS and then prove that that thing is less than my RHS. Thankful for help!
First, you have a mistaken sign here: $$e^{-x}=1-x+\frac{x^{2}}{2}-\frac{x^{3}}{6}\color{red}{+}x^{4}\frac{e^{-\Theta_2x}}{24},0\leq\Theta_2\leq1$$ Next, since the inequality is even (i.e. unchanged by the substitution $x\mapsto -x$), you can assume $x\ge 0$. The left-hand side becomes $$\frac{x^4}{24}\left | e^{\Theta_1x}+e^{-\Theta_2x} \right |=\frac{x^4}{24}\left(e^{\Theta_1x}+e^{-\Theta_2x}\right) $$ We have $$e^{\Theta_1x}\le e^1<3\\e^{-\Theta_2x}\leq e^0=1\\\Rightarrow e^{\Theta_1x}+e^{-\Theta_2x} < 4 $$ and you can conclude. Note that this also shows the expression inside the absolute value is non-negative.