Đáp án: $P_{min}=6⇔x=y=1$
Giải thích các bước giải:
Do $xy>0⇒x;y$ cùng dấu
Do vậy:
`2+\frac{1}{xy}=\frac{x^3}{2y}+\frac{y^3}{2x}+\frac{2}{x^2y^2}`
`≥2\sqrt{\frac{x^3}{2y}.\frac{y^3}{2x}}+\frac{2}{x^2y^2}`
`=2\sqrt{\frac{x^2y^2}{4}}+\frac{2}{x^2y^2}`
`=2.\frac{xy}{2}+\frac{2}{x^2y^2}=xy+\frac{2}{x^2y^2}`
`⇔2+\frac{1}{xy}-xy-\frac{2}{x^2y^2}≥0`
`⇔(2-\frac{2}{x^2y^2})+(\frac{1}{xy}-xy)≥0`
`⇔\frac{2x^2y^2-2}{x^2y^2}-\frac{x^2y^2-1}{xy}≥0`
`⇔(x^2y^2-1)(\frac{2}{x^2y^2}-\frac{1}{xy})≥0`
`⇔(xy-1)(xy+1).\frac{2-xy}{x^2y^2}≥0`
$⇔(xy-1)(2-xy)≥0$ (do `\frac{xy+1}{x^2y^2}>0`)
$⇔1≤xy≤2$
Ta có: `P=x^2y^2+x^2+y^2+1+\frac{2}{2xy-1}`
`=(x^2y^2+1)+(x^2+y^2)+\frac{2}{2xy-1}`
`≥2xy+2xy+\frac{2}{2xy-1}`
`=(4xy-2+\frac{2}{2xy-1})+2`
`≥2\sqrt{(4xy-2).\frac{2}{2xy-1}}+2`
`=2\sqrt{4}+2=6`
Dấu bằng xảy ra
$⇔\begin{cases}\dfrac{x^3}{2y}=\dfrac{y^3}{2x}\\x^2y^2=1\\x^2=y^2\\4xy-2=\dfrac{2}{2xy-1}\end{cases}⇔x=y=1$ (thỏa mãn $ĐKXĐ$)
