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The statement is

$e^x(1-e^x)\leqslant (1/4)$ for $x\lt 0$

I think the above statement is true by calculating approximate value at some points.

But how to prove it properly?

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    $\begingroup$ Hint: let $z=e^x$ $\endgroup$ – lulu Feb 9 '16 at 12:30
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Just write: $$\forall x\in\mathbb{R},\ \mathrm{e}^x(1-\mathrm{e}^x)=-\left(\mathrm{e}^x-\frac12\right)^2+\frac14.$$ Then use the fact that the square of a real number is non-negative to obtain: $$\forall x\in\mathbb{R},\ \mathrm{e}^x(1-\mathrm{e}^x)\leq\frac14.$$ (So the inequality is valid, even for non-negative $x$'s). You also get, for free, that the equality occurs exactly once, for $x=-\ln(2)$.

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  • $\begingroup$ you have made it more simpler $\endgroup$ – Mittal G Feb 10 '16 at 2:38
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Let $y=e^x$. Then $y\in (0,1)$

Can you show that for all $y\in (0,1)$, $y(1-y)\leq\frac{1}{4}$? This is just a second degree polynomial.

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Note that $e^x\in(0,1)$ for $x<0$.

Note $y(1-y)$ is a '$\bigcap$' quadratic with roots at zero and one and so by symmetry has a maximum at the midpoint $y=1/2$ where it equals $1/4$.

Restricting to $y=e^x\in (0,1)$ gives $e^x(1-e^x)\in(0,1/4]$.

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  • $\begingroup$ @ jp McCarthy thanks. got it. $\endgroup$ – Mittal G Feb 9 '16 at 12:40

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