Question

已知x> 0,y>-1,且x+y=1,求x分之1+(y+1)分之2

Answer 1

∵ x+y=1,
∴ y=-x+1
∵ y>-1
∴ -x+1＞-1
∴ -x＞-2
∴ x＜2
∵ x＞0
∴ 0＜x＜2
∵ y=-x+1
∴ 1/x+2/(y+1) = 1/x+2/(-x+1+1) = 1/x-2/(x-2)
令f(x)=1/x-2/(x-2)，
求导得：
f ′(x)=-1/x²+2/(x-2)²
= [-(x-2)²+2x²] /[x²(x-2)²]
= (x²+4x-4)/[x²(x-2)²]
= (x+2+2√2)(x+2-2√2)/[x²(x-2)²]
x属于（0，2√2-2）时，f(x)↓； x属于（2√2-2，2）时，f(x) ↑
当x=2√2-2时有最小值：
1/(2√2-2) - 2/(2√2-2-2) = 1/[2(√2-1)] + 1/(2-√2) = (√2+1)/2 + (2+√2)/(4-2) = (3+2√2)/2