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

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?

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

2 Answers 2

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\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}

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

By the first point, if we show that $$\forall x>0: x<x+1+e^{-x},$$ 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 $$\forall x>0: -1<e^{-x},$$ (just add $-x-1$ in both sides of the inequality). Then, your initial problem was reduced to proving that $\forall x>0: -1<e^{-x}$, but by our second point this is trivially true.

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  • $\begingroup$ But how did you start? I tried to prove this and I couldn't $\endgroup$ Commented Apr 23, 2021 at 18:31
  • $\begingroup$ To prove what? The fact that $ln(x)$ is a strictly increasing function? $\endgroup$
    – Amelian
    Commented Apr 23, 2021 at 18:32
  • $\begingroup$ I just don't understand what do you mean by all this. $\endgroup$ Commented Apr 23, 2021 at 18:37
  • $\begingroup$ 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 $x>0$. $\endgroup$
    – Amelian
    Commented Apr 23, 2021 at 18:41
  • $\begingroup$ 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. $\endgroup$ Commented May 3, 2021 at 15:13

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