áp dụng BĐT Cauchy :$\frac{x^{2}y^{2}}{\left[ {(4x-1)y-x} \right]^{2}}$+$\frac{(4x-1)y-x}{xy}$+$\frac{(4x-1)y-x}{xy}$$\geq$3
$\Rightarrow$$\frac{x^{2}y^{2}}{\left[ {(4x-1)y-x} \right]^{2}}$$\geq$3-$\frac{2\left[ {(4x-1)y-x} \right]}{xy}$=-5+$\frac{2}{x}$+$\frac{2}{y}$(1)
$x^{2}$+$\frac{1}{8x}$+$\frac{1}{8x}$$\geq $$\frac{3}{4}$$\Rightarrow $$x^{2}$$\geq $$\frac{3}{4}$-$\frac{1}{4x}$(2)
$y^{2}$+$\frac{1}{y}$+$\frac{1}{y}$$\geq$3$\Rightarrow$$y^{2}$$\geq$3-$\frac{2}{y}$(3)
cộng từng vế của (1)(2)(3)$\Rightarrow$P$\geq$$\frac{-5}{4}$+$\frac{7}{4x}$
do x$\geq$$\frac{1}{2}$$\Rightarrow$$\frac{7}{4x}$$\geq$$\frac{7}{2}$
$\Rightarrow$P$\geq$$\frac{9}{4}$