# Prove that $0 < g(x) - \ln x \leq \ln\left(\frac{x+2}{x}\right)$ for $x > 0$

$$g(x) = \ln(x + 1 + e^{-x})$$

My question is prove that $$0 < g(x) - \ln x \leq \ln\left(\frac{x+2}{x}\right)$$ for $$x > 0$$

How do I do that?

My attempts:

I have only successfully proved the second part:

We know that $$x > 0$$ then

$$e^{-x} + x + 1 < 2 + x$$

$$1/x$$ is positive so we multiply each other by $$1/x$$ then we put $$\ln$$ on each side too, we get this

$$\ln (\frac{e^{-x} + x + 1}{x}) < \ln\left(\frac{2 + x}{x}\right)$$

And since $$g(x) - \ln x$$ can be written as $$\ln \left(\frac{e^{-x} + x + 1}{x}\right)$$ then we proved it.

We get $$g(x) - \ln x < \ln\left(\frac{2 + x}{x}\right)$$ which is the second part.

The first part I have tried to prove it but I never succeeded:

$$x > 0$$

$$x + 1 > 1$$

$$e^{-x} < 1$$

Their signs are different so no idea what to do after this. Any idea?

• What have you tried, where are you stuck? Can you show the LHS? What about the RHS? Commented Apr 23, 2021 at 17:48
• @CalvinLin Wait a min. I'm gonna post my attempts. Commented Apr 23, 2021 at 17:48
• Do you know that $\ln(ab)=\ln a+\ln b$? Can you give a bound for $e^{-x}$ when $x>0$? Commented Apr 23, 2021 at 17:49
• @CalvinLin check the edit Commented Apr 23, 2021 at 17:58
• @HagenvonEitzen check the edit please Commented Apr 23, 2021 at 17:58

\begin{align} & \ln(x+1+e^{-x}) - \ln x \\[8pt] = {} & \ln\frac{x+1+e^{-x}} x \\[8pt] \le {} & \ln\frac{x+1+1} x \end{align}

• I don't understand. Commented Apr 23, 2021 at 18:00
• @TechnoKnight: $e^{-x}<1$ for $x>0$ Commented Apr 23, 2021 at 18:47
• @TechnoKnight : Can you be specific? Commented Apr 24, 2021 at 20:55
• @MichaelHardy I don't know what are you trying to accomplish ): Commented May 3, 2021 at 15:11
• @TechnoKnight : I'm simply proving the inequality that you gave us. Commented May 3, 2021 at 17:19

The function $$\ln(x)$$ is strictly increasing. Then, $$\ln(x) < \ln(x+1+e^{-x})$$ holds iff $$x < x+1+e^{-x}$$, i.e. iff $$-1 < e^{-x}$$, and this inequality holds for every $$x>0$$ since the exponential function is positive (indeed it holds for every real number, but here you can just consider $$x>0$$ because $$x$$ should be in the domain of the logarithmic function).

Edit (03-05-2021): Trying to explain more carefully the argument:

The facts I suppose you know about the functions $$g(z)=ln(z)$$ and $$h(z)=e^z$$ are

• The function $$g(z)=ln(z)$$ is strictly monotonically increasing; that means that $$g(z_1) < g(z_2)$$ if and only if $$z_1 < z_2$$;
• The function $$h(z)=e^z$$ can only take positive values (indeed, it's a bijection between $$R$$ and $$R^+$$).

With those points accepted, let's prove that $$\forall x>0: ln(x)

By the first point, if we show that $$\forall x>0: x we will be done, since the both statements above are equivalent (just replace $$z_1$$ by $$x$$ and $$z_2$$ by $$x+1+e^{-x}$$).

But we can rewrite our last statement as $$\forall x>0: -1 (just add $$-x-1$$ in both sides of the inequality). Then, your initial problem was reduced to proving that $$\forall x>0: -1, but by our second point this is trivially true.

• But how did you start? I tried to prove this and I couldn't Commented Apr 23, 2021 at 18:31
• To prove what? The fact that $ln(x)$ is a strictly increasing function? Commented Apr 23, 2021 at 18:32
• I just don't understand what do you mean by all this. Commented Apr 23, 2021 at 18:37
• Sorry but I don't understand what you don't understand :- ( I point out the fact that $ln(x)$ is an increasing function so we can change the problem of showing your initial inequality by the "simpler" problem of showing that $x<x+1+e^{-x}$ holds for all $x>0$. Commented Apr 23, 2021 at 18:41
• But how did you know that e^(-x) is bigger than 1? I tried proving it in my post and I found that the opposite and I couldn't complete. Can you explain your way more, please? I don't understand how ln(x) being an increasing function changes my inequality. Commented May 3, 2021 at 15:13