(19) Among the following functions, the one with a minimum value of 4 is ( ).
(A) $y=\frac{1+2 x}{\sqrt{x}}$
(B) $y=\frac{2+x^{2}}{x}$
(C) $y=\frac{1+4 x}{\sqrt{x}}$
(D) $y=\frac{4+x^{2}}{x}$
Solve $y=\frac{1+2 x}{\sqrt{x}}$ which is $y=\frac{1}{\sqrt{x}}+2 \sqrt{x}$, then $y \geqslant 2 \sqrt{2}$, when $x=\frac{1}{2}$, $y_{\text {minimum }}=2 \sqrt{2}$; $y=\frac{2+x^{2}}{x}$ which is $y=\frac{2}{x}+x$, thus, $y \geqslant 2 \sqrt{2}$ or $y \leqslant-2 \sqrt{2}$, it has no minimum value; $y=\frac{1+4 x}{\sqrt{x}}$ which is $y=\frac{1}{\sqrt{x}}+4 \sqrt{x}$, then $y \geqslant 4$, when $x=\frac{1}{4}$, $y_{\text {minimum }}=4$; $y=\frac{4+x^{2}}{x}$ which is $y=\frac{4}{x}+x$, thus, $y \geqslant 4$ or $y \leqslant-4$, it has no minimum value. Therefore, the answer is $\mathrm{C}$.