Question

若x,y属于R正,且xy-(x+y)=1,求x+y的最小值

Answer 1

x>0,y>0

则x+y≥2√xy

所以xy≤(x+y)²/4

xy-(x+y)=1

xy=x+y+1≤(x+y)²/4

令a=x+y

a+1≤a²/4

a²-4a-4≥0

a≤2-2√2,a≥2+2√2

显然a＞0

所以a≥2+2√2

所以x+y最小值=2+2√2