Verfication of deduction made using the Cauchy-Schwarz inequality

Yes, your proof is fine. Why did you write $|\sqrt{a}|$ ? If $a$ is positive, we always have $\sqrt{a}>0$.

it is $$\frac{a}{b}+\frac{b}{a}+\frac{c}{a}+\frac{a}{c}+\frac{a}{d}+\frac{d}{a}+\frac{b}{c}+\frac{c}{b}+\frac{d}{c}+\frac{c}{d}+\frac{b}{d}+\frac{d}{b}+4\geq 12+4=16$$ since $$x+\frac{1}{x}\geq 2$$ for $x>0$

While OP's proof is fine, it may be worth noting that the result also follows directly from the AM-HM inequality by simply rewriting the given relation as:

$$ \frac{a+b+c+d}{4} \;\ge\; \frac {4}{\cfrac{1}{a}+\cfrac{1}{b}+\cfrac{1}{c}+\cfrac{1}{d}} $$