I have 2 questions,
Is the following inequality true? If yes why? $$\frac{1}{2} \times ( e^{x} + e^{-x} ) \leq e^{\frac {x^2}{2}} $$
In general for real numbers $a, b$ do we have a nice upperbound like above on, $$\frac{1}{2} \times ( ae^{x} + be^{-x} )$$
Feel free to assume $a, b >0$ if needed.
$\cosh(x)\leq e^{x^2/2}$ is equivalent to $\log\cosh(x)\leq \frac{x^2}{2}$. We may safely assume $x>0$ since both sides are even.
Since $\tanh(x)\leq x$ for any $x>0$ (this is a convexity inequality) we have $$ \log\cosh(x)=\int_{0}^{x}\tanh(t)\,dt\leq \int_{0}^{x}t\,dt = \frac{x^2}{2} $$ and the first inequality is proved. Assuming that for some $a,b>0$ the inequality $2e^{x^2/2}\geq ae^x+be^{-x}$ holds, by mapping $x$ into $-x$ we have that $2e^{x^2/2}\geq be^x+ae^{-x}$ also holds, so by letting $c=\max(a,b)$ we are asking for the optimal constant $c$ such that $e^{x^2/2}\geq c\cosh(x)$ holds. This constant can be found through $$ \frac{1}{c}=\sup_{x\in\mathbb{R}} \cosh(x)e^{-x^2}, $$ but $f(x)=\cosh(x)e^{-x^2}$ is a decreasing function on $\mathbb{R}^+$, so the optimal constant is just $c=1$. Indeed $$ \log f(x) = \int_{0}^{x}\left(\tanh(t)-t\right)\,dt $$ is negative and decreasing.