<s>[INST] Prove that $\frac{\ln x}{x}+\frac{1}{e^x}<\frac{1}{2}$ for $x > 0$ 
Given $x>0$ , prove that $$\frac{\ln x}{x}+\frac{1}{e^x}<\frac{1}{2}$$

I have tried to construct $F(x)=\frac{\ln x}{x}+\frac{1}{e^x}$ and find the derivative function of $F(x)$ to find the maximun value, but I can't solve the transcendental equation.
So I tried another way.I tried to use the inequality $e^{-x}\le \frac{1}{x+1}$ (when $x>0$ ) to prove the inequality $\frac{\ln x}{x}<\frac{1}{2}-\frac{1}{e^x}$ , but I can't connect these two inequalities well, and I did't solve the problem in the end.
How can I do?
 [/INST] Update: I found a simpler proof.
It suffices to prove that
$$\frac{x}{2} - \ln x - x\mathrm{e}^{-x} > 0. $$
Let $p := \frac{23}{37} - \frac{14}{23}\mathrm{e}^{-37/23}$.
Since $p < \frac12$, it suffices to prove that
$$f(x) := px - \ln x - x\mathrm{e}^{-x} > 0. \tag{1}$$
We have
$$f'(x) = p - \frac{1}{x} + \mathrm{e}^{-x}(x-1)$$
and
$$f''(x) = \frac{1}{x^2} - (x - 2)\mathrm{e}^{-x} 
= \frac{\mathrm{e}^{-x}}{x^2}[\mathrm{e}^x - (x-2)x^2]> 0. \tag{2}$$
(Note: We have $\mathrm{e}^x - (x-2)x^2
\ge 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24}
+ \frac{x^5}{120} - (x-2)x^2 > 0$.)
Thus, $f(x)$ is convex on $x > 0$. Also, we have
$$f'\left(\frac{37}{23}\right) = 0.$$
Thus, the (global) minimum of $f(x)$ on $x > 0$ is attained at $x = \frac{37}{23}$.
Thus, we have
$$f(x) \ge f\left(\frac{37}{23}\right)
= 1 - \frac{1369}{529}\mathrm{e}^{-37/23} - \ln \frac{37}{23} > 0.$$
We are done.
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