<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] Yes, (though strangely written) your method is correct. It can be reformulated noticing more explicitely that $f(x):=4x^2-2ax+a^2-5a+4$ is decreasing for $x\le a/4$ and increasing for $x\ge a/4$, since $f'(x)=8x-2a.$ We thus recover your 3 cases (and the same numerical results as Andreas, but more directly):

*

*Case 1: $a/4\ge2$. Then, $f>0$ on $(0,2)$ iff $f(2)\ge0$, i.e. $a^2-9a+20\ge0$, i.e. $a\le2$ or $a\ge4$. This or-condition is always satisfied since $a\ge8$ by hypothesis.

*Case 2: $0<a/4<2$. Then, $f>0$ on $(0,2)$ iff $f(a/4)>0$, i.e. $\frac34a^2-5a+4>0$, i.e. $a<a_1:=\frac{10-2\sqrt{13}}3\approx0.93$ or $a>a_2:=\frac{10+2\sqrt{13}}3\approx5.74.$ This or-condition is satisfied iff $0<a<a_1$ or $a_2<a<8.$

*Case 3: $a/4\le0$. Then, $f>0$ on $(0,2)$ iff $f(0)\ge0$, i.e. $a^2-5a+4\ge0,$ i.e. $a\le1$ or $a\ge4.$ This or-condition is always satisfied since $a\le0$ by hypothesis.

Putting your three cases together gives the whole set of solutions:
$a<a_1$ or $a>a_2.$
</s>