<s>[INST] Find the range of $a$ such $|f(x)|\le\frac{1}{4},$ $|f(x+2)|\le\frac{1}{4}$ for some $x$. 
Let $$f(x)=x^2-ax+1.$$
  Find the range of all possible $a$ so that there exist $x$ with 
  $$|f(x)|\le\dfrac{1}{4},\quad |f(x+2)|\le\dfrac{1}{4}.$$

A sketch of my thoughts: I write 
$$f(x)=\left(x-\dfrac{a}{2}\right)^2+1-\dfrac{a^2}{4}\ge 1-\dfrac{a^2}{4}$$
so if $1-\dfrac{a^2}{4}>\dfrac{1}{4}$ or $-\sqrt{3}<a<\sqrt{3}$
this case impossible
But I don't know how to prove the other case, or if this there are better ideas. 
 [/INST] 
Let $$f(x)=x^2-ax+1.$$
  Find the range of all possible $a$ so that there exist $x$ with 
  $$|f(x)|\le\dfrac{1}{4},\quad |f(x+2)|\le\dfrac{1}{4}.$$

There is a solution if and only if one of the two equations $x^2-ax+1=\pm\dfrac14$ has two roots $r,s$ and $|r-s|\ge 2$ and the solution of $f(x+2)=f(x)$ satisfy $\left|f\left(\frac a2-1\right)\right|\le \frac 14.$
Part 1
$$\left|f\left(\frac a2-1\right)\right|\le \frac 14\iff 7\le a^2 \le 9.$$
Part 2
Let's work with $x^2-ax+1=\dfrac14.$ We have that $$x^2-ax+1=\dfrac 14\iff x=\dfrac{-a\pm\sqrt{a^2-3}}{2}.$$ So, there are two roots if and only if $a^2>3.$ Now
$$\left|\dfrac{a+\sqrt{a^2-3}}{2}-\dfrac{a-\sqrt{a^2-3}}{2}\right|\ge 2\iff \sqrt{a^2-3}\ge 2.$$ That is, $a^2\ge 7.$ So $a$ must satisfy three conditions: $a^2> 3,$ $a^2\ge 7$ and $7\le a^2\le 9.$ 
Part 3
If we work with $x^2-ax+1=-\dfrac14$ we get that $$x^2-ax+1=-\dfrac 14\iff x=\dfrac{-a\pm\sqrt{a^2-5}}{2}.$$ So, there are two roots if and only if $a^2>5.$ Now
$$\left|\dfrac{a+\sqrt{a^2-5}}{2}-\dfrac{a-\sqrt{a^2-5}}{2}\right|\ge 2\iff \sqrt{a^2-5}\ge 2.$$ That is, $a^2\ge 9.$ So $a$ must satisfy three conditions: $a^2> 5,$ $a^2\ge 9$ and $7\le a^2\le 9.$ 
Part 4
Thus the solution is
$$a\in [-3,-\sqrt 7]\cup [\sqrt 7,3].$$
</s>