<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 suggest a solution different from OP; albeit simpler.
First, transform the condition $\forall x\in (0,2)$. Equivalent to this condition, introduce $x = \frac{2}{y}$ with $y >1$, so only one boundary has to be inspected and the term remains quadratic in $y$ and $a$:
The inequality  $4x^2-2ax+a^2-5a+4>0$ becomes
$16-4ay+(a^2-5a+4)y^2>0$.
First note that only $a>0$ has to be inspected, since for $y > 1$ and $a \le 0$ we have
$16-4ay+(a^2-5a+4)y^2 > 16  - 9/4 + (-5/2 + a)^2 > 0$ so the condition is always obeyed for $a \le 0$.
Rewrite the question as
$$
16 - 4 a y + (a^2 - 5 a + 4) y^2 = (a y - 2 - \frac{5y}{2})^2 - \frac14 (9 y^2 + 40 y - 48) > 0 \tag{1}
$$
Since $9 y^2 + 40 y - 48 > 0$ always holds for $y >1$, we can discuss $g = (a y - 2 - \frac{5y}{2})^2$. For $a \to 0$, $g \to \frac14 (5y +4)^2 =  \frac14 (16y^2 + 64) + \frac14 (9 y^2 + 40 y - 48)$ so the question's condition is always obeyed. Likewise, for very large $a$ it is clear that (1) will also be obeyed.
Conversely, for $a=4$ we have $16 - 4 a y + (a^2 - 5 a + 4) y^2  = 16(1-y)$ which will be negative for all $y >1$.
Hence the allowed regions for $a$ will be $a_1 > a $   and $a > a_2$ with $a_2 \ge 4 \ge a_1 > 0$.
What remains is to derive $a_1$ and $a_2$ from the roots of (1) and their dependence on $y$. This is standard curve discussion and we give the results.
We have for the smaller root:
$$\frac{1}{y}\Big(2 + \frac{5y}{2} - \frac12 \sqrt{9 y^2 + 40 y - 48}\Big) 
\ge \frac{8}{5 + \sqrt{13}} = a_1 \simeq 0.93 $$
which holds since the minimum occurs at  $ y = 5 + \sqrt{13} > 1$
We have for the larger root:
$$\frac{1}{y}\Big(2 + \frac{5y}{2} + \frac12 \sqrt{9 y^2 + 40 y - 48}\Big) 
\le \frac23 (5 + \sqrt{13})  = a_2 \simeq 5.74 $$
which holds since the maximum occurs at  $ y = 5 - \sqrt{13} > 1$. $\qquad \Box$
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