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I need to prove that $ -2 + x + (2+x)e^{-x}>0 \quad \forall x>0$

If I define $f(x) = -2 + x + (2+x)e^{-x}$ and plot it I can see it's a monotonously growing function, and f(0)=0. Then $f(x)>0$ if $x>0$.

However I can't find the way to prove this. Ideally I would like to prove it without deriving the function and by using some inequalities, however I don't know if it is possible. Any hint is really appreciated.

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  • $\begingroup$ You could adapt the argument I used in this answer. $\endgroup$ Commented Oct 23, 2018 at 21:44

6 Answers 6

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We have that $f(0)=0$ and

$$f(x)=-2 + x + (2+x)e^{-x}\implies f'(x)=e^{-x}(e^x-1-x)$$

As alternative note that for $x\ge2$ the inequality is trivially satisfied then consider $0<x<2$ and we have

$$-2 + x + (2+x)e^{-x}>0 \iff e^x<\frac{2+x}{2-x}$$

and by $x=\frac2y$ with $y>1$ we obtain

$$f(y)=\left(\frac{y+1}{y-1}\right)^y=\left(1+\frac{2}{y-1}\right)^y>e^2$$

which, for my knowledge, can’t be easily proved for $y\in\mathbb{R}$ without derivatives, namely showing that $f(y)$ is monotonic.

The monotonicity of $f(y)$ can be easily proved for $y\in\mathbb{N}$ and extended to the real case assuming that $f(y)$ is convex. We could also try to extend the result to reals through rationals.

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  • $\begingroup$ <<Ideally I would like to prove it without deriving the function and by using some inequalities...>> $\endgroup$
    – rtybase
    Commented Oct 23, 2018 at 21:22
  • $\begingroup$ @rtybase Ideally means solely? I thought that it could be also useful give a solution in that classic way. $\endgroup$
    – user
    Commented Oct 23, 2018 at 21:25
  • $\begingroup$ @gimusi Are you sure that the inequality holds $\forall x>0$? I think that for $0<x<2$ it is not true. $\endgroup$
    – Iván
    Commented Oct 23, 2018 at 22:47
  • $\begingroup$ @Iván Yes indeed but for $x>2$ the original inequality is sum of positive terms. $\endgroup$
    – user
    Commented Oct 24, 2018 at 2:38
  • $\begingroup$ @gimusi sorry to insist, but I plotted the expression that you proposed and it is not true that the inequality holds. May be I am missing something, here you have a plot to check it: ibb.co/nL2BmV. I understand that for $x>2$ $f(x)>0$, but the range $0<x<2$ is the hard part to prove, and the expression that you provided (thank you by the way) I think that does not hold in that range. $\endgroup$
    – Iván
    Commented Oct 24, 2018 at 17:03
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$$ \frac{d^2}{dx^2}\left[-2+x+(2+x)e^{-x}\right] = x e^{-x} $$ hence $f(x)=-2+x+(2+x)e^{-x}$ is a convex function on $\mathbb{R}^+$.
Since $f'(0)=f(0)=0$, $f(x)$ is increasing and positive on $\mathbb{R}^+$.

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A variation on Christian Blatter's solution:

$e^{-x} (2+x) +x-2 = e^{-x} ( 1+ \frac{x-2}{x+2}) $

As $e^{-x} \gt \ 0 \ \forall \ x \in \mathbb{R} $, it's enough to check if, $(1+\frac{x-2}{x+2}) \gt 0 $ for $x\gt 0$

However, now we have: $1+\frac{x-2}{x+2} = 1+\frac{x-2+2-2}{x+2} = 2-\frac{4}{x+2}$ which is strictly increasing for $x\gt 0$ and bounded below by $ 0 = e^{-x} ( 1+ \frac{x-2}{x+2}$ ) for $x=0$

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We have to prove that $$e^{-x}>{2-x\over 2+x}\qquad(x>0)\ .\tag{1}$$ This is obvious when $x\geq2$. For $0<x<2$ replace the claim $(1)$ by $$e^x<{2+x\over2-x}=1+{x\over 1-{x\over2}}\qquad(0<x<2)\ .$$This means that we have to prove $$1+x+{x^2\over2!}+{x^3\over3!}+{x^4\over4!}+\ldots<1+x+{x^2\over2}+{x^3\over2^2}+{x^4\over 2^3}+\ldots\qquad(0<x<2) ,$$ which follows immediately by comparing terms.

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  • $\begingroup$ Just to clarify, although some other answers seems to valid as well, I found this the most elegant and simple. Thus, I am going to accept this one as the correct one. $\endgroup$
    – Iván
    Commented Oct 26, 2018 at 17:06
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This approach requires an extra condition $x\lt2.$

\begin{align}\ln\left(1+\dfrac{x}{2}\right)-\ln\left(1-\dfrac{x}{2}\right)&=\left[\dfrac{x}{2}-\dfrac{1}{2}\left(\dfrac{x}{2}\right)^2+\dfrac{1}{3}\left(\dfrac{x}{2}\right)^3+\cdots\right]-\left[-\dfrac{x}{2}-\dfrac{1}{2}\left(\dfrac{x}{2}\right)^2-\dfrac{1}{3}\left(\dfrac{x}{2}\right)^3-\cdots\right]\\&=x+\dfrac{2}{3}\left(\dfrac{x}{2}\right)^3+\cdots\\&\gt x\quad\forall x\lt2\cdots(*)\\\end{align}

$\forall x\gt0,\dfrac{2+x}{2-x}\gt1\implies\ln\left(\dfrac{2+x}{2-x}\right)\gt0$

$\therefore0\lt x\lt\ln\left(\dfrac{2+x}{2-x}\right)\implies e^{-x}\gt\dfrac{2-x}{2+x}\implies -2+x+(2+x)e^{-x}\gt0$

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  • $\begingroup$ That's also a nice way, of course you are assuming $0<x<2$ but maybe it should be specified. I was also trying to find a method which doesn't make use of derivative, convexity, integrals or series concept but only algebraic methods. $\endgroup$
    – user
    Commented Oct 25, 2018 at 10:37
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Another alternative that I found but is not as elegant as the answer marked as the accepted one is expanding the exponential and rearranging the terms as follows:

$ \left(e^{x } (x -2 )+2 +x \right) >0 \\ \iff (1+x+\frac{x^2}{2!}++\frac{x^3}{3!}+\dots) (x -2 )+2 +x >0 \\ \iff \sum_{n=3}^{\infty} x^n(\frac{1}{(n-1)!}-\frac{2}{(n-1)!n})>0 $

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