已知x,y∈(-√3,√3),且xy=-1 ,求s=3/(3-x^2)+12/(12-y^2)的最小值
1个回答

s = 3/(3-x²) + 12/(12-y²)

= [3(12-y²)+12(3-x²)] / [(3-x²)(12-y²)]

= (72-12x²-3y²)/(36-12x²-3y²+x²y²)

∵ xy = -1

∴ x²y² = 1

s = (72-12x²-3y²)/(37-12x²-3y²)

= 1 + 35/(37-12x²-3y²)

∵ 12x²+3y² ≥ 2√(36x²y²) = 12

∴ s ≥ 1 + 35/(37-12) = 12/5

当且仅当 12x² = 3y² ,即 x=1/√2,y=-√2 或 x=-1/√2,y=√2 时取得最小值.