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Is there a good upper bound for $x(1-x)$ if $xin [0,1]$?

Depends on what you mean by "good". A lot of times, $1$ is a perfectly useful upper bound for the expression. If you need a stricter upper bound, the maximum of the expression (and therefore the smallest upper bound possible) is $frac14$.

You can easily see this is the upper bound by calculating the derivative and setting it to $0$, or by rewriting the expression (completing the square) as

$$x(1-x) = x-x^2 = - left(x-frac12right)^2 + frac14$$

An upper bound is given by $frac{1}{4}$.

Define $f(x) = x(1 - x) = x - x^{2}$. Then, $f'(x) = 1 - 2x$, and we have a critical point at $x = frac{1}{2}$. Plugging this value back into our original equation yields $frac{1}{4}$.

We have $x(1-x) ge 0$ and by AM-GM

$$x(1-x)le left(frac{x+1-x}{2}right)^2=frac14$$

with equality for

• $x=1-x implies x=frac12 $

therefore as $x in[0,1]$

$$0le x(1-x)le frac14$$

Let $f(x)=x(1-x)$. $f'(x)=(1-x)+x=0$, we get $x=dfrac{1}{2}$ as the critical point. To be sure of the nature of the function (whether concave up or down) use second derivative test to confirm $f''(x)<0$. This says that the function $f$ at $x=dfrac{1}{2}$ attains a local maximum thereby giving us the upper bound $fBig(dfrac{1}{2}Big)$.