Đk : x > $\frac{1}{2}$ Ta có VT = x + (2-$\frac{1}{x}$) +y + (2-$\frac{1}{y}$)<=> VT= x+ $\frac{2x-1}{x}$ + y + $\frac{2y-1}{y}$<=> VT $\geq$ 2$\sqrt{2x-1}$ + 2$\sqrt{2y-1}$Dấu = xảy ra <=> $x^{2}$=2x-1 $y^{2}$=2y-1 <=> x=y=1 (TM)
Đk : x > $\frac{1}{2}$ Ta có VT = x + (2-$\frac{1}{x}$) +y + (2-$\frac{1}{y}$)<=> VT= x+ $\frac{2x-1}{x}$ + y + $\frac{2y-1}{y}<=> VT $\geq$ 2$\sqrt{2x-1}$ + 2$\sqrt{2y-1}$Dấu = xảy ra <=> $x^{2}$=2x-1 $y^{2}$=2y-1 <=> x=y=1 (TM)
Đk : x > $\frac{1}{2}$ Ta có VT = x + (2-$\frac{1}{x}$) +y + (2-$\frac{1}{y}$)<=> VT= x+ $\frac{2x-1}{x}$ + y + $\frac{2y-1}{y}
$<=> VT $\geq$ 2$\sqrt{2x-1}$ + 2$\sqrt{2y-1}$Dấu = xảy ra <=> $x^{2}$=2x-1 $y^{2}$=2y-1 <=> x=y=1 (TM)