jmc

By Cauchy-Schwarz, $$(x+y+z)\left(\frac{4}{x}+\frac{9}{y}+\frac{16}{z}\right)\ge (2+3+4{)}^2=81,$$so $$\frac{4}{x}+\frac{9}{y}+\frac{16}{z}\ge \frac{81}{3}=27.$$Equality occurs when $\frac{x^2}{4}=\frac{y^2}{9}=\frac{z^2}{16}.$ Along with the condition $x+y+z=3,$ we can solve to get $x=\frac{2}{3},$ $y=1,$ and $z=\frac{4}{3},$ so the minimum value is $\boxed{27}.$