2x+8y-xy=0 2x+8y=xy 2/y + 8/x=1 x+y = 2x+8y-xy=0 2x+8y=xy 2/y + 8/x=1 x+y =(x+y)*1 =(x+y)(2/y + 8/x) =8+2+ 2x/y +8y/x ≥10+2√[(2x/y)(8y/x)] =10+2√16=18 The minimum value is 18 What you don't understand = 8 + 2 + 2x / y + 8y / X ≥10+2√[(2x/y)(8y/x)] =10+2√16=18 How can 2x / y + 8y / X be equal to 2 √ [(2x / y) (8y / x)] and 2 √ 16

2x/y +8y/x>=2√[(2x/y)(8y/x)]=2√16
Used
a+b.>=2√(ab )

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