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Prove or disprove: $\sqrt{x-x^2} \leq \frac{1}{2}$ for $x \in [0,1]$?

I have to prove this inequality since I have seen the figure of the parabola and it is very clear that for $x \in [0,1]$ it holds. But where do I begin? Can someone give me a hint?

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Hint:$$x-x^2=\frac14-\left(x-\frac12\right)^2.$$

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  • $\begingroup$ So $x-x^2 \leq \frac{1}{4}$ and then just take square roots? Thanks a lot! $\endgroup$ – mandella Jun 12 '18 at 7:04
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    $\begingroup$ @mandella Indeed. $\endgroup$ – José Carlos Santos Jun 12 '18 at 8:00
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$$\dfrac{x+1-x}2\ge\sqrt{x(1-x)}$$

for $x\ge0,1-x\ge0\iff0\le x\le1$

Alternatively, $x=\sin^2t,0\le t\le\dfrac\pi2$

$$\implies\sqrt{x(1-x)}=\dfrac{\sin2t}2$$

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1) Expression under square root must be $\ge 0.$

Indeed, considering $x(1-x) \ge 0 \rightarrow $

$x \in [0,1]$ (Why?).

Squaring :$\sqrt{x-x^2} \le 1/2$ gives

$x-x^2 \le 1/4$, or

$x^2-x +1/4 \ge 0$, or

$(x-1/2)^2 \ge 0.$

Hence ?

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