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We have to prove the following inequalities:

1) to show that $\frac{2x}{\pi }<sin\left(x\right)<x,\:and\:after\:1-e^{-\frac{\pi }{2}}\le \int _0^{\frac{\pi }{2}}\:e^{-sin\left(x\right)}dx<\frac{\pi }{2}\left(1-e^{-1}\right)$

2) to show that $e^{-1}\le e^{-x^2}\le \frac{1}{1+x^2}\:and\:after\:\frac{1}{e}\le \int _0^1\:e^{-x^2}dx\le \int _0^1\:\frac{1}{1+x^2}$

For 2) I almost done: for $x\in \left[0,1\right]$ we have $x^2\le 1$ which involving $\:e^{-x^2}\ge e^{-1}$ and after integration we obtain that $\frac{1}{e}\le \int _0^1\:e^{-x^2}dx$, but I don't know how to show that $e^{-x^2}\le \frac{1}{1+x^2}$, because if I prove that, after integration I obtain what I want... so for second remain just to prove $e^{-x^2}\le \frac{1}{1+x^2}$ and after don't make problem to finish.

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Let $g(x)=1-(1+x^2)e^{-x^2}$. Then $g'(x)=2x^3e^{-x^2}\geq 0, \forall x\in [0,1].$ So $g$ is a an increasing function and $g(x)\geq g(0)=0$. Thus $1-(1+x^2)e^{-x^2} \geq 0$ wich is equivalent to $\frac{1}{1+x^2}\geq e^{-x^2}$.

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    $\begingroup$ But how did you realize which is the function g to use derivative ? $\endgroup$ – Andrei Mihai Apr 17 '15 at 8:35
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    $\begingroup$ Because $\frac{1}{1+x^2}\geq e^{-x^2} \Leftrightarrow \frac{1}{1+x^2}-e^{-x^2}\geq 0 \Leftrightarrow 1-(1+x^2)e^{-x^2}\geq 0$. So you have to solve $g(x)\geq 0$. $\endgroup$ – Patissot Apr 17 '15 at 8:40
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    $\begingroup$ have an idee for 1) ? $\endgroup$ – Andrei Mihai Apr 17 '15 at 10:04
  • $\begingroup$ The first inequality is classical, you can prove it by using the same ideas, study $f(x)=\sin (x)-x$ and $g(x)=\sin (x) -\frac{2}{\pi}x$. You can also use convexity. $\endgroup$ – Patissot Apr 17 '15 at 10:51
  • $\begingroup$ Just I did that, but it seem very easiest with classical method... another way is not possible? ps: what you mean with convexity? can you show me at answers? $\endgroup$ – Andrei Mihai Apr 17 '15 at 10:54
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$e^{x^2}=1+x^2$ when $x=0$. However since $\frac{d}{dx}e^{x^2}=2xe^{x^2}$ and $\frac{d}{dx} {1+x^2}=2x$, $\frac{d}{dx}e^{x^2}>\frac{d}{dx} {1+x^2}$ for all $x>0$. Hence, $e^{x^2}>{1+x^2}$ for all $x>0$, hence $e^{-x^2}<\frac{1}{1+x^2}$ for all $x>0$.

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