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

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 
$$-\frac1{12} \le y^2-y+\frac 16\le \frac16.$$
