<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] Let $g(x)=Ax^2+Bx+C$,if $A>0$, then $g(x)\ge 0$
Case 1: $\forall x \in \Re$, if $B^2\le 4AC$.
Case 2: For $x \in [p,q]$, if the position of the min (bottom) of $g(x)$, $x_0=-\frac{B}{2A}$ should be s.t $x_0 \le p$ or $x_0\ge q$.
Here $$f(x)=4x^2-2ax+a^2-5a+4>0$$ in $x\in (0,2)$ can be met in both the cases.
Case 1: $4a^2 \le 16(a^2-5a+4) \implies 3a^2-20a+16 \ge 0 \implies (x-x_1)(x-x_2)\ge 0$ where $x_1=\frac{2}{3}(5-\sqrt{13}) \approx 0.92), x_2=\frac{2}{3}(5+\sqrt{13}) \approx 5.73$. So $$a\le \frac{2}{3}(5-\sqrt{13})~ \text{or}~ a\ge \frac{2}{3}(5+\sqrt{13}).....(1)$$
Case 2: $x_0=a/4$, $x_0<0 \implies a\le 0$  and $x_0\ge 2 \implies a\ge 8.$
So in this case $$a\le 0 ~\text{or}~ a\ge 8........(2)$$.
Interestingly, the final domain of $a$ needs to be the union of (1) and (2).
Lastly, $f(x)> 0, ~\text{ONLY}~\forall x \in(0,2)$, when $a\in (-\infty,0)\cup (8,\infty).$
Otherwise,  $f(x)>0, \forall x \in \Re $, when $a \in (-\infty,\frac{2}{3}(5-\sqrt{13}) \cup  (\frac{2}{3}(5+\sqrt{13}), \infty)$
</s>