Show that $|\{x\}^2-\{x\}+1/6|\leq \frac{1}{6}$

Question

I am trying to show the following inequality holds for $x>0$: $$|{x}^2-{x}+1/6|\leq \frac{1}{6}.$.

I was able to show that $\{x\}^2-\{x\}+1/6 \leq 1/6$ because $\{x\}^2-\{x\}<0.$ However I am having trouble with showing $\{x\}^2-\{x\}+1/6\geq -\frac{1}{6}$. This seems to be untrue because this would mean $\{x\}^2-\{x\}\geq -\frac{1}{3}$ which seems impossible if $x$ is chosen just right. But am not able to find a counterexample either.

Could anyone please help me with the second half of the proof above. I appreciate the help!

Answer 1

You need to prove $\{x\}^2-\{x\} + \frac{1}{3} \geq 0$

But $a^2 - a + 1/3 \gt 0$ for all real values of $a$

A) Just find the discriminant $D = 1^2 - 4\cdot1\cdot \frac{1}{3}$.
It is negative. This proves that $a^2 - a + 1/3 \gt 0$ for all $a$.

Or...

B) Alternatively, just write this:

$a^2 - a + 1/3 = (a^2 - a + 1/4) + 1/12 = (a-1/2)^2 + 1/12 \gt 0$

So... that proves it.

Side note: "which seems impossible if $x$ is chosen just right"
Well, this is a typical example how our intuition can lie to us.

Answer 2

Hint: Find the solutions to the inequality $$t^2-t\geq -\frac 13 $$ It turns out they are all $t\in\mathbb{R}$.

Answer 3

Set $y=\{x\}$. As $x>0$, we have $0\le y <1$, so, as the roots of $y^2-y$ are $0$ and $1$, its minimum is attained at $\frac12$, and we have $-\frac14\le y^2 -y\le 0\:$ on $[0,1)$. Therefore $$-\frac1{12} \le y^2-y+\frac 16\le \frac16.$$