$\exists C>0$ such that $\frac{1}{2}(e^x+e^{-x}) \le e^{\frac{1}{2}x^2-Cx}$?

Question

It is well-known and easy to check that for any real $x$ it holds $$ \frac{1}{2}(e^x+e^{-x}) \le e^{\frac{1}{2}x^2}. $$ [To show this, it is sufficient to write explicitly their Taylor series expansion]

Question: does there exists a positive constant $C$ for which for all reals $x$ it holds $$ \frac{1}{2}(e^x+e^{-x}) \le e^{\frac{1}{2}x^2-Cx}\,\,? $$

In other words, can we improve the upper bound on the hyperbolic cosine?

Answer 1

The left hand side is $1+o(x)$ the right and side is $1-C.x+o(x)$ so the difference is $-C.x+o(x)$ and is $<0$ if $x$ is small $>0$

Answer 2

Well, I made the calculations, and $$ e^{\frac{1}{2}x^2}-\frac{1}{2}\left(e^x+e^{-x}\right)=\frac{x^4}{4}+O(x^6), $$ so..