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So I am learning this chapter "mean value theorem", and there is an exercise.

Prove the inequality of $e^{x} \gt x+1$ for $x$ different from zero and the inequality of $2x \arctan x \ge \ln (x^2+1)$... I feel very retarded every time I read this exercise because I don't understand what theorem should I use?

Can you help me just a little and I'll do the rest by myself?

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Hey, don't feel "retarded" man, it is just a matter of experience. After a little while, you will be solving these like butter and bread. –  Anon Jan 31 '13 at 21:03
    
For the one with $e^x$, how about checking the Taylor series for $e^x$? –  Anon Jan 31 '13 at 21:04
    
homework should not be used as a standalone tag; see tag-wiki and meta. –  Martin Sleziak Feb 5 '13 at 9:27

2 Answers 2

For establishing the inequality $e^x>1+x$, consider two cases: $x>0$ and $x<0$.

For the case $x>0$, apply the Mean Value Theorem to the function $f(x)=e^x$ over the interval $[0,x]$. This gives a $c$ with $0<c<x$ satisfying $$ e^x-e^0 =(x-0)\cdot e^c=xe^c. $$ But $e^0=1$ and $e^c>1$ (strict inequality, since $c>0$). so $$ e^x-1>x; $$ which implies the result for $x>0$.

I'll leave the other case for you...


I'll just give a hint for your second inequality:

For the second inequality, break things up into two cases: $x\ge 0$ and $x<0$.

For the case $x\ge0$, apply the Mean Value Theorem to the function $g(x)=2x\arctan x -\ln(x^2+1)$ over the interval $[0,x]$ and use the fact that $\arctan(c)\ge 0$ for all $c\ge0$. (Note: the second case is easier here, since $g$ is an even function.)

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I wouldn't use the mean value theorem there. For the first one:

$e^x>x+1$ since $y=x+1$ is the tangent line to the graph $y=e^x$ in the point $(0,1)$. Since $e^x$ is convex it's graph is above the tangent line.

For the second one I would define $$ f(x) = 2x \arctan x - \log(x^2+1) $$ and study the sign of the derivative $f'(x)$ to prove that $0$ is a global minimum.

addendum Just to see how you get a rigorous proof.

Consider $f(x)=e^x-x-1$. Notice that $f'(x) = e^x-1$. For $x> 0$ we have $f'(x)>0$ hence $f(x)$ is increasing in $[0,+\infty)$. For $x<0$ we have $f'(x)<0$ hence $f(x)$ is decreasing on $(-\infty,0]$. Hence for all $x\neq 0$ we have $f(x) > f(0) = 0$. This means $e^x > x+1$.

For the second one take $g(x) = 2x\arctan x - \log(x^2+1)$. We have $g'(x) = 2\arctan x$ and the proofs is the same as before.

edit So now I realize what you mean by "mean value theorem". In fact to prove that $f'>0$ implies $f$ increasing you use that theorem.

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But rigorous proofs of things like those rely on the mean value theorem. –  Michael Hardy Jan 31 '13 at 21:08
    
My proofs are rigorous :-) –  Emanuele Paolini Jan 31 '13 at 21:08
    
"Convex" is defined by saying the secant line segments lie above the graph. How does one prove that if the second derivative is positive, then the function is convex, and if it's convex then it lies above its tangent lines? The answer is that one uses the mean value theorem to do that. –  Michael Hardy Jan 31 '13 at 21:10
    
@manu-fatto Since you can easily edit your answer, I suggest not using the same letter ($f$) for both functions. –  Git Gud Jan 31 '13 at 21:17
    
@MichaelHardy: You are right! What you (anglophone) mean with "mean value theorem" is not what I thought. In Italy this is called "Lagrange Theorem" –  Emanuele Paolini Jan 31 '13 at 21:19

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