Proof: For all real numbers $x$, If $x^2−5x+4≥0$, then either $x≤1$ or $x≥4$. Is this a valid technique? For all real numbers $x$, If $x^2−5x+4≥0$, then either $x≤1$ or $x≥4$. Is this a valid technique?
Instead of showing that either (x-4) and (x-1) are both positive or both negative, can I just do what I have written below instead?
$x^2−5x+4=(x-4)(x-1)≥0$
There are two possible situations: $x≥4$ or $x<4$.
if $x<4$ then $(x-4)<0$ and so we get $(x-1)≤ 0$ which implies that $x≤1$.
so either $x≤1$ or $x≥4$.

Yes, it is valid. Well done!
Here I propose an alternative way to solve it:
\begin{align*}
x^{2} - 5x + 4 = \left(x^{2} - 5x + \frac{25}{4}\right) - \frac{25}{4} + 4 = \left(x - \frac{5}{2}\right)^{2} - \frac{9}{4} \geq 0
\end{align*}
which is equivalent to
\begin{align*}
\left|x - \frac{5}{2}\right|^{2} \geq \frac{9}{4} \Longleftrightarrow\left(x - \frac{5}{2} \geq \frac{3}{2}\right)\vee\left(x - \frac{5}{2} \leq -\frac{3}{2}\right)\Longleftrightarrow (x \geq 4)\vee(x\leq 1)
\end{align*}
Hopefully this helps.
