a) Ta có: $3\geq \frac{1}{x}+\frac{4}{y}\geq \frac{(1+2)^2}{x+y}=3$
Do đó: dấu '=' xảy ra $\Leftrightarrow \frac{1}{x}=\frac{2}{y}\Leftrightarrow 2x=y$; mà $x+y=3\Rightarrow x=1;y=2$
b) Ta có: $3=1/x+4/y+9/z\geq (1+2+3)^2/(x+y+z)\geq 36/12=3$
Do đó: dấu '=' xảy ra $\Leftrightarrow 1/x=2/y=3/z; x+y+z=12\Rightarrow x=2;y=4;z=6$