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This question already has an answer here:

The problem is to prove the following inequality:

$$ (e^x - 1) \ln(1+x) > x^2 , \quad\text{ for } x >0 $$

Let me introduce notation $f(x) > g(x)$.

At $x=0$ both sides are equal to $0$. So, to prove the inequality one could prove it not for original functions, but for their derivatives. Unfortunately, derivatives of the functions doesn't seem to simplify the situation at all.

I've also tried to substitute some parts of the original inequality with the new variable like this:

  • $ z = e^x $;
  • $ z = \ln(1+x) $.

Finally, I've tried to separate $e^x$ and $\ln(1+x)$ (which is the major obstacle for good differentiation) by dividing both sides of the inequality by one of these functions.

None of the above helped me to move significantly towards proof.

I'm kindly asking for your help. Both hints and complete solutions are very appreciated.

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marked as duplicate by Daniel Fischer May 3 '16 at 12:50

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ sometimes you can write as $f(x) > 0$ and prove $f`(x)>0, \, x>0$ and usually $f(0)=0$, hence one has $f(x)>f(0)=0, \, x>0$. $\endgroup$ – Chip May 3 '16 at 7:38
  • $\begingroup$ I've mentioned differentiation in the question. Subtracting $x^2$ doesn't really help - the resulting expression is as hard to prove as the original one. $\endgroup$ – soupault May 3 '16 at 7:47
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With $f(x)=(\operatorname e^x-1)\ln(1+x)$ and $g(x)=x^2$ it is easy to show that $f(0)=g(0)$ and $f'(0)=g'(0)$ since all of these are equal to zero. Digging deeper we find that $f''(0)=g''(0)=2$, and finally we reach $f'''(0)=g'''(0)=0$. Now note that $$ f'''(x)=(\operatorname e^x-1)\cdot\frac{2}{(1+x)^3}+\underbrace{3\operatorname e^x\cdot\frac{-1}{(1+x)^2}+3\operatorname e^x\cdot\frac1{1+x}}_{\text{greater than zero}}+\operatorname e^x\ln(1+x)>0 $$ for $x>0$, since we see that the negative contribution from the second term is clearly compensated by the positive contribution from the third term. The first and last terms above are clearly positive for $x>0$.

So working backwards we have $f'''(x)>g'''(x)$ implies $f''(x)>g''(x)$ implies $f'(x)>g'(x)$ implies $f(x)>g(x)$ for $x>0$.


Note that the principle for higher order derivatives of a product $p(x)q(x)$ is given as $$ \begin{array}{ccc} (p(x)q(x))'&=&p'(x)q(x)+p(x)q'(x)\\ (p(x)q(x))''&=&p''(x)q(x)+2p'(x)q'(x)+p(x)q''(x)\\ (p(x)q(x))'''&=&p'''(x)q(x)+3p''(x)q'(x)+3p'(x)q''(x)+p(x)q'''(x)\\ &\text{etc.} \end{array} $$ where the coefficients $1\ 1$ in the first row, $1\ 2\ 1$ in the second, and $1\ 3\ 3\ 1$ in the third row are the renowned binomial coefficients. So with $p(x)=(\operatorname e^x-1)$ and $f(x)=\ln(1+x)$ this enables us to compute derivatives of $f(x)=p(x)q(x)$ efficiently.

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  • $\begingroup$ how can one prove $f"(x)>2, \, x>0$? $\endgroup$ – Chip May 3 '16 at 8:59
  • $\begingroup$ @Chip: Maybe not as elegant as you hoped for, but nonetheless details have been added. $\endgroup$ – String May 3 '16 at 9:41
  • $\begingroup$ many thanks! Side add-on: the derivatives of a product of functions follow the Newton's binom with powers replaced by derivative orders. $\endgroup$ – Chip May 3 '16 at 9:58
  • $\begingroup$ @Chip: OK, thank you - I will make a last edit to get rid of superfluous details. $\endgroup$ – String May 3 '16 at 10:06
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    $\begingroup$ @String: thanks! Since this question seems to be close, I can't write another possible solution (I had this idea but I got stuck since I didn't prove $f^{(3)}(x)>0$). So, I will put it here: if one takes $f(x)$ and writes its Taylor expansion around $0$, the term $x^2$ shows up and from the expression of the reminder $\propto f^{(3)}(\xi) x^3>0 $ ($\xi>0$), hence $f(x)-x^2>0$. $\endgroup$ – Chip May 4 '16 at 1:36
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Let $A(x,e^x-1)$, $B(x,-\ln(1+x)$ and $O(0,0)$.

Now easy to see that $\measuredangle AOB>90^{\circ}$ and we are done!

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  • $\begingroup$ You left all the "connecting the dots" to the reader, but interesting perspecitve on it nonetheless. Since $f(x)=\operatorname e^x-1>x$ and since $g(x)=-\ln(1+x)$ has the same graph as $f(x)$ but rotated $90^\circ$ clockwise we see that indeed $\angle BOA>90^\circ$ for $x>0$. The dot product then reveals that $$x\cdot x+(\operatorname e^x-1)(-\ln(1+x))<0$$ and after rearranging the result follows. $\endgroup$ – String May 3 '16 at 12:48
  • $\begingroup$ Wow! That's a great solution! I wonder if one could occasionally come with it or this method is typical for wide variety of problems. Thanks for explanation, @String! $\endgroup$ – soupault May 3 '16 at 16:10
  • $\begingroup$ Nice take! thanks everyone here for an interesting problem and comments... $\endgroup$ – Chip May 4 '16 at 1:44
  • $\begingroup$ PS: I have doubts about the proof of Michael: if it relies on the fact that $-\ln(1+x) < -x$ (so $-\ln(1+x)$ lies below the second bisector), that inequality is not true. It would remain to show that the angle is obtuse by other means, if that holds at all (!?). $\endgroup$ – Chip May 4 '16 at 2:25
  • $\begingroup$ @Chip: Sorry, my explanation was unclear. Since 1. $f(x)=(\operatorname e^x-1)>x$ and since 2. $f$ is convex, we see 1. that $y=f(x)>x=f(z)>z=ln(1+x)$ so therefore $A=(x,y)$ has traversed more of the graph of $f$ than $B=(x,-z)$ has traversed of the rotated graph of $f$. Thus by 2. $OA$ is tilted by a greater angle counterclockwise than $OB$. It follows that the angle between them must be obtuse. $\endgroup$ – String May 5 '16 at 22:23
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$e^x-1\geq x+\dfrac{x^2}{2}+\dfrac{x^3}{6}$ and $\ln(1+x)\geq\dfrac{x}{1+x}$ for $x>0$.

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