If $x$ is a real and $0 < x < 4$, then $\frac{4}{x(4-x)} \geq 1$ I got this exercise from the textbook Book of Proof, CH4 E12. I've tackle this problem in the following manner:
Suppose x is a real and $0 < x < 4$, it follows that,
\begin{align*}
&\Rightarrow 0 - 2 < x - 2 < 4 - 2 \\
&\Rightarrow     4 < (x - 2)^2 < 4\\
&\Rightarrow     0 \leq (x - 2)^2 < 4
\end{align*}
Since, $x(4 - x) = 4x - x^2 = 4 - (x - 2)^2$, then
$$\dfrac{4}{x(4 - x)} = \dfrac{4}{4 - (x - 2)^2}.$$
This expression is greater or equal to $1$ for
$0 \leq (x - 2)^2 < 4$. Thus,
$$\dfrac{4}{x(4 - x)} \geq 1.$$
I'm quite new to proof technique and I'm using this book to self-learn logic and proofing writing. My question is: is the solution stated above logically sound? Would my arguments be considered sufficient to prove that $P \Rightarrow Q$?

We can do this proof by contradiction: We want to prove $P\rightarrow Q$ and we prove it by assuming $P$ and $\neg Q$, and getting a contradiction.
Suppose $0<x<4$ with $x\in\mathbb{R}$, and the inequality $\dfrac{4}{x(x-4)}\geq 1$ does not hold. 
Then we have $0<x,0<4-x$ (both follow from $0<x<4$) and $\dfrac{4}{x(x-4)}<1$, and we have:
\begin{align*}
\dfrac{4}{x(4-x)}&<1 & &\\
4&<x(4-x) & &\text{[multiplying both sides by $x(x-4)>0$]}\\
0&<x(4-x)-4\\
0&<-x^2+4x-4\\
x^2-4x+4&<0\\
(x-2)^2&<0
\end{align*}
which is absurd because every squared real number is positive. 
This contradiction allows us to conclude that $\dfrac{4}{(x-4)}\geq 1$.
