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Show that $e^{-x} \sin x < \frac{x}{1+x},x>0$.

Trial: Let $f(x)=e^{-x} \sin x - \frac{x}{1+x}$ So, $f'(x)=e^{-x}(\cos x- \sin x)-\frac{1}{(1+x)^2}$. From here I can't conclude anything. I also think about series of $e^{-x},\sin x , (1+x)^{-1}$.But I am unable to solve.Please help.

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up vote 6 down vote accepted

Hint: Using this inequality: If $x>0$, then $$e^x > 1+x$$ and $$\sin x < x$$

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thanks. Great answer. – Argha Jan 2 '13 at 4:29
(+1) nice hint. – Mhenni Benghorbal Jan 2 '13 at 4:36
+1 Nice. Just a little thing: multiplying the inequalities works fine as long as $\sin x >0\,$ , otherwise...well, otherwise the wanted inequality is trivial, yet some little care is required here. – DonAntonio Jan 2 '13 at 5:01

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