<s>[INST] Find the all possible values of $a$, such that $4x^2-2ax+a^2-5a+4>0$ holds $\forall x\in (0,2)$ 
Problem: Find the all possible values of $a$, such that
$$4x^2-2ax+a^2-5a+4>0$$
holds $\forall x\in (0,2)$.


My work:
First, I rewrote the given inequality as follows:
$$
\begin{aligned}f(x)&=\left(2x-\frac a2\right)^2+\frac {3a^2}{4}-5a+4>0\end{aligned}
$$
Then, we have
$$
\begin{aligned}
0<x<2\\
\implies -\frac a2<2x-\frac a2<4-\frac a2\end{aligned}
$$
Case $-1:\,\,\,a≥0 \wedge 4-\frac a2≤0$.
This leads,
$$
\begin{aligned}\frac {a^2}{4}>\left(2x-\frac a2\right)^2>\left(4-\frac a2\right)^2\\
\implies \frac {a^2}{4}+\frac {3a^2}{4}-5a+4>f(x)>\left(4-\frac a2\right)^2+\frac {3a^2}{4}-5a+4\end{aligned}
$$
For $f(x)>0$, it is enough to take $\left(4-\frac a2\right)^2+\frac {3a^2}{4}-5a+4>0$ with the restriction $a≥0\wedge 4-\frac a2≤0$.
Case $-2:\,\,\,a≤0 \wedge 4-\frac a2≥0$.
We have:
$$
\begin{aligned}\frac {a^2}{4}<\left(2x-\frac a2\right)^2<\left(4-\frac a2\right)^2\\
\implies \frac {a^2}{4}+\frac {3a^2}{4}-5a+4<f(x)<\left(4-\frac a2\right)^2+\frac {3a^2}{4}-5a+4\end{aligned}
$$
Similarly, for $f(x)>0$, it is enough to take $\frac{a^2}{4}+\frac {3a^2}{4}-5a+4>0$ with the restriction $a≤0\wedge 4-\frac a2≥0$.
Case $-3:\,\,\,a≥0 \wedge 4-\frac a2≥0$.
This case implies, $\left(2x-\frac a2\right)^2≥0$.  This means, $f(x)≥\frac {3a^2}{4}-5a+4$
Thus, for $f(x)>0$, it is enough to take $\frac {3a^2}{4}-5a+4>0$ with the restriction $a≥0\wedge 4-\frac a2≥0$.
Finally, we have to combine all the solution sets we get.

I haven't done the calculation, because I want to make sure that the method I use is correct. Do you see any flaws in the method?
 [/INST] I found the solution sets in your cases:
Case-1) $((-\infty,4)\cup(5,\infty))\cap[8,\infty)=[8,\infty)$
Case-2) $((-\infty,1)\cup(4,\infty))\cap[-\infty,0)=(-\infty,0]$
Case-3) $((-\infty,\frac{10-2\sqrt{13}}{3})\cup(\frac{10+2\sqrt{13}}{3},\infty))\cap[0,8]=[0,\frac{10-2\sqrt{13}}{3})\cup(\frac{10+2\sqrt{13}}{3},8]$
The union of sets we found in all cases is $S=(-\infty,\frac{10-2\sqrt{13}}{3})\cup(\frac{10+2\sqrt{13}}{3},\infty)$.  S is the solution of the problem.
My solution:
$f'(x)=0$ gives the minumum of the quadratic function $f(x)=4x^2-2ax+a^2-5a+4$ at $x=\frac{a}{4}$ and the solution of the inequality $f(\frac{a}{4})=\frac{3}{4}a^2-5a+4>0$ is the set $S$ above.
If $a$ were in $[\frac{10-2\sqrt{13}}{3},\frac{10+2\sqrt{13}}{3}]$ then $x=\frac{a}{4}$ would be in $(0,2)$ and $f(\frac{a}{4})$ would be non-positive. So, we can conclude that $S$ is the solution set of the problem.
</s>