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!

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.  
