what's an elegant way to show that $x(1-x) \leq \frac14$? for $x \in \mathbb{R}$, consider $f(x) = x(1-x)$, using traditional methods of finding global extremas, we can show that the derivative has a unique zero at $x= \frac12$ and $f''(\frac12) < 0$, thus $x(1-x) \leq \frac14 = f(\frac12)$
is there a more elegant way ?
 A: Notice that $(1/2-x)^2 \geq 0$.  The statement is trivial to prove from this.
A: Since it’s a quadratic polynomial with two zeros, the extreme value is at the vertex, whose $x$ coordinate is midway between the zeros of the polynomial. 
So all you have to do is find the value $x=\frac 12$ midway between the zeros, then compute $f(\frac12)$ and confirm it is positive and therefore a maximum. 
A: The interesting case is when $0<x<1$ (the others are obvious). For this case use AM-GM inequality for the numbers $x$ and $1-x$:
$$\sqrt{x(1-x)}\leq \frac{x+(1-x)}{2}$$
A: Easy to see that all the values outside $I = (0,1)$ will be negative, so the optimal solution must lie in $I$. Now you are optimizing an area of a rectangle with fixed perimeter of $1$, which is known from geometry to be a square, so $x= 1-x$ which implies $x=1/2$.
A: If $x<0$ or $x>1$ the expression is negative, so we need only consider $0\leq x\leq 1$, in which case the function is symmetric around $x=1/2$, increasing to the left of $1/2$ and decreasing to the right of it, hence maximised at $x=1/2$, where the value is $1/4$.
A: 1) x >1; inequality is trivial(x(1-x)<0).
2) x <0; inequality is trivial(x(1-x)<0).
Consider $0\le x \le 1$.
Set $x = \sin^2y$ , $0\le y \le π/2$.
Then 
$\sin^2 y(1-\sin^2 y)=\sin^2 y \cos^2 y=$
$(1/4)\sin^2 2y \le 1/4.$
A: Yes, this is a quadratic function, so it has an extrem at $p={x_1+x_2\over 2} = {0+1\over 2}$ and this extreme is $f(p)= f(1/2)=1/4$. 
Since the leading coefficient is negative this extrem is a maximum.
A: Let $y = x-1/2$, so we have $-(y + 1/2)(y - 1/2) = -y^2 + 1/4$, which is clearly $\leq 1/4$.
