Why is it true that if $x \in [0,1], x(1-x) \le 1/4$

Question

I am reading through M. Nair's classic proof about the lower bound of the Least Common Multiple when I encountered this step:

for $x \in [0,1], x(1-x) \le 1/4$

While I can see this is true by testing it out with any value of $x$, I am not able to see the argument for why it is always true.

Could someone state the argument for why this inequality holds?

Answer 1

Assuming that you meant $\leqslant$ instead of $<$, you can do it as follows:\begin{align}x(1-x)\leqslant\frac14&\iff x-x^2\leqslant\frac14\\&\iff0\leqslant x^2-x+\frac14\\&\iff\left(x-\frac12\right)^2\geqslant0.\end{align}

Answer 2

With a bit more (calculus) machinery, let $f(x)=x(1-x)$. The extreme value theorem implies that $f$ has a maximum on $[0, 1]$. Either the maximum occurs at one of the endpoints $f(0)=f(1)=0$ or in the interior in which case the derivative $f'(x)=1-2x$ must vanish. We note that $f'(1/2)=0$ and $f(1/2)=1/4> f(0)=f(1)=0$. Thus for all $x\in [0,1]$, $$ f(x)=x(1-x)\leq 1/4 $$ with equality iff $x=1/2$.

Answer 3

By AM-GM inequality you have for all $x \in [0,1]$ $$\sqrt{x(1-x)} \leq \frac{x+1-x}{2}$$ with equality if and only if $x=\frac{1}{2}$.

squaring you get $$x(1-x) \leq \frac{1}{4}$$ with equality when $x=\frac{1}{2}$ so no strict inequality.

p.s. If $x \notin [0,1]$ then exactly one of $x$ or $1-x$ is positive, meaning $x(1-x) <0 <\frac{1}{4}$, so the inequality actually holds for all $x$.

Answer 4

We have that

$$x(1-x)\le c \iff x^2-x+c\ge 0$$

and

$$\Delta=1-4c=0 \implies c=\frac14$$

Answer 5

Let $x = \frac 12 + e$ and then $1- x = \frac 12 - e$. Then $x(1-x) = (\frac 12 +e)(\frac 12 - e) = \frac 14 - e^2$. As $e^2 \ge 0$ it is true that $x(1-x) \le \frac 14$ but it is not true that $x(1-x) < \frac 14$.

If $x = \frac 12$ then $x(1-x) = \frac 14$. But if $x \ne \frac 12$ and $e \ne 0$ then $x(1-x) = \frac 14 - e^2 < \frac 14$.

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There's many way to view it but basically $x(1-x) \le \frac 14 \iff -x^2 + x -\frac 14 \le 0$. $-x^2+x-\frac 14$ is a parabola that "opens downward" and has a maximum.

You can use any formula for parabola's to find that if $f(x) = ax^2 + bx + c$ will have axis of symmetry and max/min value at ... $x = -\frac b{2a}$ and the max/min value is .... $a(-\frac b{2a})^2 + b(\frac b{2a}) + c$....

But these all boil down to $ax^2 + bx + c = a(x + \frac b{2a})^2 + (c-\frac {b^2}{4a^2})$ which has max/min value when $(x + \frac b{2a})^2 = 0$

In this case $-x^2 + x= -(x - \frac 12)^2 + \frac 14$. This has maximum value when $x-\frac 12=0$. And that max value is $\frac 14$.

.....

Which is basically how we go about thinking about the AM-GM theorem in which $x > 0, 1-x > 0$ so $\frac {x + (1-x)}2 \ge \sqrt{x(1-x}$ or $\frac 12 \ge \sqrt{x(1-x}$ and $\frac 14 \ge x(1-x)$.

Answer 6

$x(1-x)$ is a quadratic polynomial, hence it has a global extremum, which is a maximum because its leading coefficient is negative.

On the other hand, the high school theory of quadratic equations shows the extremum of the quadratic function $ax^2+bx+c$ is attained at $x=-\frac b{2a}$, which is also the arithmetic mean of its roots (real or complex). In the present case the roots are $0$ and $1$, so the maximum is attained at $x=\frac12$.