Từ \frac{1}{x}+\frac{1}{y}-\frac{2}{z}=0\Rightarrow z=\frac{2xy}{x+y}.Thay vào ta có P=\frac{x+\frac{2xy}{x+y}}{2x-\frac{2xy}{x+y}}+\frac{y+\frac{2xy}{x+y}}{2y-\frac{2xy}{x+y}}=\frac{x+3y}{2x}+\frac{3x+y}{2y}=\frac{1}{2}+\frac{3y}{2x}+\frac{3x}{2y}+\frac{1}{2}\geqslant1+3/2 (Áp dụng BĐT cauchy)...
Từ
$\frac{1}{x}+\frac{1}{y}-\frac{2}{z}=0\Rightarrow z=\frac{2xy}{x+y}
$.Thay vào ta có
$P=\frac{x+\frac{2xy}{x+y}}{2x-\frac{2xy}{x+y}}+\frac{y+\frac{2xy}{x+y}}{2y-\frac{2xy}{x+y}}=\frac{x+3y}{2x}+\frac{3x+y}{2y}
$ $=\frac{1}{2}+\frac{3y}{2x}+\frac{3x}{2y}+\frac{1}{2}\geqslant1+
\frac{3
}{2}.2
=4$ (Áp dụng BĐT cauchy)...