Why is $\frac{1}{n}x(1-x)\leq \frac{1}{4n}$ for $x\in[0,1]$? Why is $\frac{1}{n}x(1-x)\leq \frac{1}{4n}$ for $x\in[0,1]$?
I don't see why this is true, just from looking at the left hand side of the inequality.
 A: Why don't you phrase the question by asking why $x(1-x)\le \dfrac 1 4$?
Here's one way to see it:
\begin{align}
x(1-x) & = -(x^2 -x) \\[10pt]
&  = \frac 1 4 -\left( x^2-x+\frac 1 4 \right) \tag{completing the square} \\[10pt]
& = \frac 1 4 -\left( x - \frac 1 2  \right)^2.
\end{align}
This is $1/4$ minus a square; hence $1/4$ minus a positive number; hence less than $1/4$, UNLESS $x=1/2$, in which case the square vanishes and it's exactly $1/4$.
A: We can set $x=\sin^2y\implies x(1-x)=\dfrac{\sin^22y}4=\dfrac14-\dfrac{\cos^22y}4\le\dfrac14$
A: As $0\le x\le1,0\ge-x\ge-1,0\le 1-x\le1$
Using AM $\ge GM,$
$$\frac{x+1-x}2\ge\sqrt{x(1-x)}$$
A: For $$f(x) = x(1-x) = x -x^2$$ we get $$f'(x) = 1-2x= 0 \iff x = \frac{1}{2}$$ so the maximum of $f$ can only be reached for $x \in \{0,1/2,1\}$ and
$$f(0)=f(1)=0, \quad f\left(\frac{1}{2}\right) = \frac{1}{4},$$
it follows that $f(x) \leq 1/4$ for every $x \in [0,1]$. 
A: One way you can see this is by maximizing $x(1-x)$:
$$x(1-x) = x - x^2 \Rightarrow \dfrac{d}{dx}(x-x^2) = 1-2x$$
Setting this equal to $0$ and solving for $x$ yields $x=1/2$, which is indeed a maximum on the interval $[0,1]$. (For $f(x) = x(1-x),~f''(1/2) = -2 < 0$). 
Hence, $x(1-x) \leq 1/4$ on the interval $[0,1] \Rightarrow \dfrac{1}{n} \cdot x(1-x) \leq \dfrac{1}{4n}$ on the interval $[0,1]$. 
Sorry if that notation doesn't make sense.... my first time every replying to something on StackExchange and I haven't ever messed around with HTML.
