Analysis

This question examines the application of basic inequalities to find the minimum value. Paying attention to when the equality holds is key to solving the problem, making it a basic question.

By applying basic inequalities to find the minimum value, we can verify each option in turn.

Solution

For option A, it can be transformed into $y=\sqrt{{x}^{2}+1}+ \dfrac{1}{ \sqrt{{x}^{2}+1}} \geqslant 2$, which holds if and only if $\sqrt{{x}^{2}+1}= \dfrac{1}{ \sqrt{{x}^{2}+1}}$, i.e., when $x^2=0$. Therefore, this option is correct.

For option B, it can be transformed into $y=x-1+ \dfrac{1}{x-1}+2$, which is increasing in $(3,+\infty)$, so $y > \dfrac{9}{2}$, hence incorrect.

For option C, when $0 2$, making it incorrect.

For option D, since $x$ can be positive or negative, we cannot conclude that the minimum value equals $2$, hence incorrect.

Therefore, the correct choice is $\boxed{A}$.