# 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?

• The maximum of about $0.4957$ occurs when $x \approx 1.617579$ so you do not have a lot of room to play with Commented Nov 10, 2022 at 14:23
• Thank you. I used GeoGebra to get the same result, but I can't write such a such an answer on the homework book. I need nigorous proof, so I came to the forum to ask this question. Commented Nov 10, 2022 at 14:46
• Incidentally, the opening phrase is usually worded something like "Given $x > 0$, prove that..." Commented Nov 11, 2022 at 3:51
• Thank you very much! This is the first time I have asked math questions in English. Commented Nov 12, 2022 at 3:19

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.

• This proof uses little numerical approximation (with the exception of the calculation of $g(b)$ at the end) and feels systematic and elegant to me. Commented Nov 13, 2022 at 13:52
• @L.F. Thanks. My answer is still complicated. Commented Nov 13, 2022 at 14:29
• I update my proof. Commented Nov 13, 2022 at 15:29
• Very ingenious shift of the extreme point. I wish I could upvote twice. Commented Nov 13, 2022 at 19:16
• @L.F. Thanks again. Commented Nov 13, 2022 at 22:57

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 $$[1.7, +\infty)$$.

$$F(x)=\frac{\ln x}{x}+e^{-x} \qquad \implies \qquad F'(x)=\frac{1-\log (x)}{x^2}-e^{-x}$$

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

$$\left( \begin{array}{ccc} n & x_1 & F(x_1) &\text{method}\\ 2 & 1.617082135 & 0.495693411 & \text{Newton} \\ 3 & 1.617582181 & 0.495693446 & \text{Halley} \\ 4 & 1.617579113 & 0.495693446 & \text{Householder} \\ 5 & 1.617579115 & 0.495693446 & \text{no name} \\ \cdots &\cdots & \cdots &\cdots \\ \infty & 1.617579115 & 0.495693446 & \text{exact} \\ \end{array} \right)$$