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. 

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 $g(x)=\frac{x^2}{2}$.
