<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 think your current description is not satisfactory.
Take Case 3 for example. You said,

This means, $f(x) \ge \frac {3a^2}{4}-5a+4$. Thus, for $f(x) > 0$, it
is enough to take $\frac {3a^2}{4}-5a+4>0$ with the restriction $a \ge
 0\wedge 4-\frac a2\ge 0$.

This description does not emphasize the sufficient and necessary condition to include all possible values of $a$.
It looks (at least from the description) like a sufficient condition (more precisely, $\frac {3a^2}{4}-5a+4>0$ is a sufficient condition of $f(x) > 0$).
One way to overcome this is given below (Cases 1-2 are omitted here, and given later):
Let
$$f(x) := 4x^2 - 2ax + a^2 - 5a + 4 = \left(2x-\frac a2\right)^2+\frac {3a^2}{4}-5a+4. \tag{1}$$
We have
$$0 < x < 2 \implies -\frac a2<2x-\frac a2<4-\frac a2. \tag{2}$$
We split into three cases.
Case 3: $-\frac{a}2 < 0$ and $4 - \frac{a}2 > 0$, i.e. $0 < a < 8$
Note that $\frac{a}{4} \in (0, 2)$
and $f(\frac{a}{4}) = \frac {3a^2}{4}-5a+4$.
We have,
\begin{align*}
 &f(x) > 0, \quad \forall 0 < x < 2\\
 \iff \quad & \frac{3a^2}{4}-5a+4 > 0\\
 \iff \quad & a > \frac{10}{3} + \frac23\sqrt{13}, \quad \mathrm{or} \quad a < \frac{10}{3} - \frac23\sqrt{13}.
\end{align*}
Thus, the set of qualified $a$ is $0 < a < \frac{10}{3} - \frac23\sqrt{13}$
or $\frac{10}{3} + \frac23\sqrt{13} < a < 8$.

Someone may ask why I took Case 3 for example, rather than Case 1 or 2? The reason is that the conditions in Case 1 and Case 2 (respectively, $a \ge 8$ and $a \le 0$), actually both imply $f(x) > 0$. By contrast, the condition of Case 3 (i.e. $0 < a < 8$) does not imply $f(x) > 0$.
The details are given below.
Case 1: $4 - \frac{a}{2} \le 0$, i.e. $a \ge 8$
From (1) and (2), we have
\begin{align*}
 &0 < x < 2 \\
 \implies\quad & \left(2x-\frac a2\right)^2 > \left(4 - \frac{a}2\right)^2\\
 \implies \quad & f(x) > \left(4 - \frac{a}2\right)^2 + \frac {3a^2}{4}-5a+4 = (a - 4)(a - 5) > 0.
\end{align*}
Thus, the set of qualified $a$ is $a \ge 8$.
Case 2: $-\frac{a}2 \ge 0$, i.e. $a \le 0$
From (1) and (2), we have
\begin{align*}
 &0 < x < 2 \\
 \implies\quad & \left(2x-\frac a2\right)^2 > \left( - \frac{a}2\right)^2\\
 \implies \quad & f(x) > \left(- \frac{a}2\right)^2 + \frac {3a^2}{4}-5a+4 = (a - 1)(a - 4) > 0.
\end{align*}
Thus, the set of qualified $a$ is $a \le 0$.
Summarily, all possible values of qualified $a$ is $a < \frac{10}{3} - \frac23\sqrt{13}$
or $a > \frac{10}{3} + \frac23\sqrt{13}$.
</s>