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I need to prove that the following inequality holds:

$\int_{0}^{1} \sqrt{x}\space e^{-x^2}dx \leq \frac{\pi}{6}$

No progress on it, yet. Any suggestion is welcome. Thanks.

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    @MichaelBoratko: Seems to be true, my bad.2012-06-04

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If Cauchy-Schwarz inequality is in one's toolkit, one can write $ \left(\int_0^1\sqrt{x}\mathrm e^{-x^2}\mathrm dx\right)^2\leqslant\left(\int_0^1x\mathrm e^{-2x^2}\mathrm dx\right)\cdot\left(\int_0^1\mathrm dx\right)=\left[-\tfrac14\mathrm e^{-2x^2}\right]_0^1=\tfrac14(1-\mathrm e^{-2}). $ Hence $ I=\int_0^1\sqrt{x}\mathrm e^{-x^2}\mathrm dx\leqslant\tfrac12\sqrt{1-\mathrm e^{-2}}\lt\tfrac12\lt\tfrac\pi6. $ Edit: The numerical approximation of $I$ above is not so bad since the bound $\mathrm e^{-x^2}\geqslant\mathrm e^{-x\sqrt{x}}$ for every $x$ in $(0,1)$ yields the lower bound $I\geqslant\frac23(1-\mathrm e^{-1})\approx0.4214$, to be compared with the upper bound $\tfrac12\sqrt{1-\mathrm e^{-2}}\approx0.4619$ (while the appearance of $\frac\pi6\approx0.5236$ in the picture remains a mystery to me is convincingly explained by @Chris in a comment below).

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    Using the fact that $e^x \ge 1 + x$ we have that $\int_{0}^{1} \sqrt{x}\space e^{-x^2}dx \leq \int_{0}^{1} \frac{\sqrt{x}}{1+x^2}dx \leq \int_{0}^{1} \frac{\sqrt{x}}{1+x^3}dx=\frac{\pi}{6}$.2012-07-02