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I'm trying to proof the following inequality: $x+1<e^{x}<2x+1$ for $0<x\leq log2$
I used the MVT on with $f(t)=e^{t}$ on the interval $(0,x)$ to proof that $x+1<e^{x}$ and I tried to do the same with $(x, log2)$ but I only got to $e^{x}<2x-2log2+2$.
Any help or hint will be appreciated.

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  • $\begingroup$ If you have shown that $e^x < 2x+2-2\log(2)$ then use $2- 2\log(2) < 1$ to prove the upper limit. $\endgroup$ – Math Lover Aug 15 '17 at 20:56
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Consider the function $g(t)=e^t-(2t+1)$. We see that $g(0)=0$. Assume that there exists $c\in(0,\log 2)$ such that $g(c)=0$. Now, apply Rolle's theorem (which is only a special case of MVT). You should arrive at a contradiction quickly.

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Hint:

The left-hand side inequality is true for all $x$ and results from $\mathrm e^x$ being a convex function.

As to the right-hand side inequality apply it on $[0,x]$: you obtain a $y\in (0,x)$ such that $\;\mathrm e^x-1=x\mathrm e^y$. Can you bound $\mathrm e^y$ independently of $x$, knowing $x\in (0,\log 2)$?

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The Taylor series at the origin of $\frac{e^x-1}{x}$ (which is an entire function) only has positive coefficients, hence it is trivial that $f(x)=\frac{e^x-1}{x}$ is both increasing and convex on $[0,\log 2]$. In particular

$$ \forall x\in(0,\log 2),\quad f(0)< f(x) < f(\log 2) $$ which is equivalent to $$ \forall x\in(0,\log 2),\quad x+1 < e^x < \frac{x}{\log 2}+1. $$ $\frac{1}{\log 2}<2$ is equivalent to $e<4$.

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