<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 want to bring the most straightforward way to calculate this problem systematically. Also, you can check the attached graph for better representation. I write the quadratic equation $f_x(a)=4x^2-2ax+a^2-5a+4$ w.r.t. $a$ as an unknown parameter. We have:
$f_x(a)=a^2-(2x+5)a+(4x^2+4)>0$.
So, $a$ must be greater than the bigger root of $f_x(a)=0$ or less than the smaller root, which are:
$f_x(a)=0 \rightarrow a=x+2.5\pm\sqrt{-3x^2+5x+2.25}=g^\pm(x)$ for $x\in(0,2)$.
So, we have $a>g^+(x)$ or $a<g^-(x)$ for $\forall x \in (0,2)$, i.e.:
$a>\max_{x\in(0,2)}g^+(x)$ or $a<\min_{x\in(0,2)}g^-(x)$. We have:
$dg^+(x)/dx=0\rightarrow 1+\frac{-6x+5}{2\sqrt{-3x^2+5x+2.25}}=0\rightarrow -12x^2+20x+9=36x^2-60x+25$
$\rightarrow 3x^{2}-5x+1=0\rightarrow x=\frac{5+\sqrt{13}}{6}\rightarrow g^+(\frac{5+\sqrt{13}}{6})=\frac{2}{3}(5+\sqrt{13})$
Similarly, we have:
$dg^-(x)/dx=0\rightarrow 1-\frac{-6x+5}{2\sqrt{-3x^2+5x+2.25}}=0 \rightarrow x=\frac{5-\sqrt{13}}{6} \rightarrow g^-(\frac{8}{5+\sqrt{13}})=\frac{8}{5+\sqrt{13}}$
So, $a$ range having $f_x(a)=4x^2-2ax+a^2-5a+4>0$ are:
$a>\frac{2}{3}(5+\sqrt{13}) \approx5.7370$ or $a<\frac{8}{5+\sqrt{13}} \approx0.9296$

</s>