Đáp án:
`(x^2+y^2)/(x-y)`
`=(x^2-2xy+y^2+2xy)/(x-y)`
`=((x-y)^2+2)/(x-y)`
`=(x-y)+2/(x-y)`
Do `x>y=>x-y>0`
`<=>2/(x-y)>0`
Nên ta áp dụng bất đẳng thức cosi với hai số dương ta có:
`(x-y)+2/(x-y)>=2\sqrt{(x-y)*2/(x-y)}=2\sqrt{2}`
Dấu "=" xảy ra khi:\(\begin{cases}xy=1\\x-y=\dfrac{2}{x-y}\\x>y\\\end{cases}\)
`<=>` \(\begin{cases}xy=1\\(x-y)^2=2\\\\\end{cases}\)
`<=>` \(\begin{cases}xy=1\\x-y=\sqrt{2}(do\,x>y)\\\end{cases}\)
`<=>` \(\begin{cases}x=y+\sqrt{2}\\y(y+\sqrt{2})=1\\\end{cases}\)
`<=>` \(\begin{cases}x=y+\sqrt{2}\\y^2+\sqrt{2}y=1\\\end{cases}\)
`<=>` \(\begin{cases}x=y+\sqrt{2}\\y^2+2.y.\dfrac{\sqrt{2}}{2}+\dfrac{1}{2}=\dfrac{3}{2}\\\end{cases}\)
`<=>` \(\begin{cases}x=y+\sqrt{2}\\\left(y+\dfrac{\sqrt{2}}{2}\right)^2=\dfrac{3}{2}=\dfrac{6}{4}\\\end{cases}\)
`<=>` \(\left[ \begin{array}{l}\begin{cases}y=\dfrac{\sqrt{6}-\sqrt{2}}{2}\\x=y+\sqrt{2}=\dfrac{\sqrt{6}-\sqrt{2}+2\sqrt{2}}{2}=\dfrac{\sqrt{6}+\sqrt{2}}{2}\\\end{cases}\\\begin{cases}y=\dfrac{-\sqrt{6}-\sqrt{2}}{2}\\x=y+\sqrt{2}=\dfrac{-\sqrt{6}-\sqrt{2}+2\sqrt{2}}{2}=\dfrac{-\sqrt{6}+\sqrt{2}}{2}\\\end{cases}\end{array} \right.\)
