prove or disprove the inequality $\sqrt{x-x^2} \leq \frac{1}{2}$ for $x \in [0,1]$

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?

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

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

$$\dfrac{x+1-x}2\ge\sqrt{x(1-x)}$$

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

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

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

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 ?