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?
 A: Let
\begin{align*}
f(x) &= e^{-x} + \frac{\log (x)}{x}, \\
f'(x) &= -e^{-x} + \frac{1 - \log (x)}{x^2}, \\
f''(x) &= e^{-x} + \frac{2 \log (x) - 3}{x^3}.
\end{align*}
We would like to show that $f(x) < 1/2$ for all $x > 0$.
Case 1. $0 < x \le 1$
Using the inequalities $e^x > x + 1$ and $\log(x) \le x - 1$,
$$ \begin{split}
\frac{x e^x}{2} (1 - 2 f(x))
&= \frac{1}{2} e^x (x-2 \log (x))-x \\
&> \frac{1}{2} (x + 1) (x - 2(x - 1)) - x \\
&= - \frac{1}{2} (x-1) (x+2) \\
&< 0.
\end{split} $$
Therefore, $f(x) < 1/2$.
Case 2. $1 < x \le 1.7$
We have $2 \log (x) - 3 < 0$.  Hence,
$$ \begin{split}
f''(x) &= e^{-x} + \frac{2 \log (x) - 3}{x^3} \\
&< e^{-1} + \frac{2 \log(1.7) - 3}{(1.7)^3}
\approx -0.0267 < 0.
\end{split} $$
(Such approximations can obviously be made rigorous by using interval arithmetic.)
Thus, $f'$ is strictly decreasing on $[1, 1.7]$.
Since
$$
f'(1) = 1 - \frac{1}{e} \approx 0.632
$$
and
$$
f'(1.7) = -e^{-1.7} + \frac{1 - \log(1.7)}{1.7} \approx -0.020,
$$
there is a unique $\alpha \in (1, 1.7)$ such that
$$ f'(\alpha) = -e^{-\alpha} + \frac{1 - \log (\alpha)}{\alpha^2} = 0 $$
and $f$ reaches its maximum on $[1, 1.7]$ at $\alpha$.  It can then be shown via interval arithmetic that $\alpha \approx 1.618$ and $f(\alpha) \approx 0.496 < 1/2$.
Case 3. $x > 1.7$
Let
$$
g(x) = x^2 f'(x) = -x^2 e^{-x} + 1 - \log(x).
$$
Then,
$$
g'(x) = \frac{e^{-x} \left(x^3 - 2x^2 - e^x\right)}{x}.
$$
It is left as an exercise for the reader to show that $x^3 - 2x^2 - e^x < 0$, so $g'(x) < 0$ for all $x > 0$.  But $g(1.7) < 0$, so for all $x > 1.7$, $g(x) < 0$ and thus $f'(x) < 0$.  Therefore, $f$ is decreasing on $[1.7, +\infty)$.
A: 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.
A: $$F(x)=\frac{\ln x}{x}+e^{-x} \qquad \implies \qquad F'(x)=\frac{1-\log (x)}{x^2}-e^{-x}$$
Instead of looking for the zero of $F'(x)$, it is better to consider
$$g(x)=1-\log(x)-x^2 e^{-x}$$
Obviously, the root is very close to $x=\sqrt e$ since $g(\sqrt e)=\frac{1}{2}-e^{1-\sqrt{e}}=-0.0227$.
Using this as an $x_0$ for a Newton like method of order $n$, we have explicit formulae for $x_1$. For example, the first iterate of Newton method is
$$\color{blue}{x_1=\sqrt e-\frac 12\,\frac{\sqrt{e}
   \left(2e^{1-\sqrt{e}}-1\right)}{1+\left(2-\sqrt{e}\right)
   e^{1-\sqrt{e}}}}$$
Now, the results
$$\left(
\begin{array}{ccc}
n & x_1 & F(x_1) &\text{method}\\
 2 & 1.617082135 & 0.495693411 & \text{Newton} \\
 3 & 1.617582181 & 0.495693446 & \text{Halley} \\
 4 & 1.617579113 & 0.495693446 & \text{Householder} \\
 5 & 1.617579115 & 0.495693446 & \text{no name} \\
\cdots &\cdots &  \cdots &\cdots   \\
 \infty & 1.617579115 & 0.495693446 & \text{exact} \\
\end{array}
\right)$$
