Proving $\forall x\in\mathbb R : \dfrac{e^x + e^{-x}}2 \le e^{\frac{x^2}{2}}$ with Cauchy's MVT 
Prove for all $x\in\mathbb R$:
  $$\dfrac{e^x + e^{-x}}2 \le e^{\frac{x^2}{2}}$$

Mclauren expansion:
$$e^x=1+x+\frac{x^2}{2}+\frac{x^3}{3!}+R_4(x)$$
$$e^{-x}=1+-x+\frac{x^2}{2}-\frac{x^3}{3!}+S_4(x)$$
Adding both together: $2+\frac{x^2}{2}+R_4(x)+S_4(x)$
Question: Can I cancel $R_4(x)+S_4(x)$ because they tend to $0$ ? if yes then this is how I continued: 
We have now: $2+\dfrac{x^2}2 \le e^{\frac{x^2}{2}}$, or $\dfrac{2+\dfrac{x^2}2}{e^{\frac{x^2}{2}}} \le 1$.
Now I thought about using Cauchy's MVT here: define: $f(x)=2+\dfrac{x^2}2 , \ g(x)= e^{\frac{x^2}{2}}$ both are continous and differentiable on all $\mathbb R$ so: 
$$\frac {f(x)} {g(x)}=\frac {f'(x)-f'(0)}{g'(x)-g'(0)}=\frac{1}{e^{\frac{x^2}{2}}}\le1$$
This is related to: Proving that $\frac{e^x + e^{-x}}2 \le e^{x^2/2}$ but my approach is different and I want to know if it's correct.
Note: I can't use integration. 
 A: An alternative approach:
$$ \ln\cosh x = \int_0^x \tanh t\,dt \le \int_0^x t\,dt = \frac{x^2}{2} $$
This uses the fact that $\tanh x\le x$ for $x\ge 0$ (and the reverse for $x\le 0$), which can be proved by computing the second derivative of $\tanh$, concluding that it's convex on $(\infty,0]$ and concave on $[0,\infty)$, and comparing it to its tangent line at $x=0$.  Another way is to use the fact that if $f$ is convex on $[a,b]$ then
$$ \frac1{b-a}\int_a^b f(x)\,dx
= \int_0^1 f((1-t)a+tb) \,dt
\le \int_0^1 \big((1-t)f(a)+tf(b)\big)\,dt
= \frac{f(a)+f(b)}{2} $$
Applying this to $f(x)=e^x$ on $[-u,u]$ yields $\frac{\sinh u}{u}\le\cosh u$, which gives the needed inequalities for $\tanh$.
Edit.  You've added a note to the question saying that you can't use integration.  The first line of this answer can be reformulated to accommodate that restriction: prove that if $f(a)\le g(a)$ and $f'(x)\le g'(x)$ for all $x\in[a,b]$, then $f(x)\le g(x)$ for all $x\in[a,b]$ (this is a consequence of MVT); then take $f(x)=\ln\cosh x$ and $g(x)=\frac{x^2}{2}$.
A: \begin{align}
\frac{e^x+e^{-x}}{2} & = 1+\frac{x^2}{2} + \frac{x^4}{24} + \frac{x^6}{720} + \frac{x^8}{40230} + \cdots \\[12pt]
e^{x^2/2} & = 1 + \frac{x^2}{2} + \frac{x^4}{8} + \frac{x^6}{48} + \frac{x^8}{384} + \cdots
\end{align}
If you can show that the denominators must remain smaller in the second series than in the first, that does it.
$$
(2n)! > n! 2^n \text{ for }n\ge 2 \text{ ?}
$$
That should be provable by induction.  In the sequence $\{(2n)!\}_{n=0,1,2,3,\ldots}$, when $n$ is incremented to $n+1$, one multiplies by $(2n+1)(2n+2)$.  In the sequence $\{n!2^n\}_{n=0,1,2,3,\ldots}$, one multiplies by $2(n+1)$.  In the induction step, all you need is that $2(n+1)<(2n+1)(2n+2)$ for $n\ge 2$.
PS: If you add your two Maclaurin series, you get $2+x^2 + R_4(x)+S_4(x)$. (You have $x^2/2$ where you need $x^2$.)  The ratios of $R_4(x)$ and $S_4(x)$ to $x^3$ both approach $0$ as $x\to0$.  From that it follows that the ratio of their sum to $x^3$ also approaches $0$.  That doesn't say whether their sum is positive, negative, or $0$.
You say "We now have $\text{[something]}\le e^{x^2/2}$".  That is not correct.  "We now have" normally means "We have now proved".  This isn't something you've proved; it's something it would now suffice to prove (if what preceded it had been correct).
In your proposed application of Cauchy's mean value theorem, you have the roles of $f$ and $f'$ incorrectly interchanged.  You also seem to be using $x$ to refer to two different things.  You need to look closely at what the theorem says before you try to apply it.
The question is really about what comes after the second-degree terms in the power series.  If you look at the difference
$$
e^{x^2/2} - \frac{e^x+e^{-x}}{2},
$$
you would expand it as a power series in which the first non-zero term is a fourth-degree term.  You need to show that that is always positive.
