the inequality $\ln^2 x \le 4e^{-2} (x + x^{-1}) \quad (x>0)$ 
For all $x>0$ prove the inequality:  $$\ln^2 x \le 4e^{-2} \left(x + x^{-1}\right).$$

To prove this I tried straightforward differentiation with $f(x) = 4e^{-2} (x + x^{-1}) - \ln^2 x$. The critical points satisfy a transcendental equation (a numerical plot indicates that there are three of these points and that the inequality is not quite sharp). Is there a better way to rewrite the original statement before attempting differentiation? Or perhaps I should abandon differentiation as an approach here?
 A: With $x=e^u>0$ we see
\begin{align}
\cosh u =&\sqrt{1+\sinh^2u}\\
\geq&1+\frac12\sinh^2u\\
=&\frac18\left(6+e^{2u}+e^{-2u}\right)\\
\geq&\frac18\left(8+4u^2+\frac43u^4\right)\\
\geq&\frac18e^2u^2
\end{align}
A: Hint: $$u = \ln(x) \Rightarrow u^2 \leq 4e^{-2}(e^u + e^{-u})$$
Which $u \in (-\infty, +\infty)$.
A: HINT: defining $$f(x)=\frac{4}{e^2}\left(x+\frac{1}{x}\right)-(\ln(x))^2$$
then the first derivative is given by $$f'(x)=4\,{\frac {1}{{{\rm e}^{2}}} \left( 1-{x}^{-2} \right) }-2\,{\frac {
\ln  \left( x \right) }{x}}
$$ show that $x=1$ is the only solution of $$f'(x)=0$$
A: Since our inequality does not depend on the substitution $x\rightarrow\frac{1}{x}$, we can assume $x\geq1$ 
and we need to prove that
$$2\sqrt{x+\frac{1}{x}}\geq e\ln{x}.$$
We'll prove a stronger inequality: $g(x)\geq0$, where
$$g(x)=2\sqrt{x+\frac{1}{x}}-\sqrt{\frac{1119008}{150735}}\ln{x}.$$
Indeed, $g'(x)=0$ gives 
$$(39x-5)(5x-39)(773x^2+390x+773)=0,$$
which gives $x_{min}=7.8$ and since $g(7.8)=0.03466...>0$, we are done!
