最小值为5/7
函数化简为f(x)=1-2/(ax+1)(a大于0,且a不等于1)
f(x1)+f(x2)=2-2/(ax1+1)-2/(ax2+1)=1
化成整式得到
a²x1x2-a(x1+x2)=3
x1,x2,是正实数
根据不等式性质有a²x1x2≤a²(x1+x2)²/4
所以a²(x1+x2)²/4-a(x1+x2)≥3整理得到
a²(x1+x2)²-4a(x1+x2)-12≥0
[a(x1+x2)-6][a(x1+x2)+2]≥0(因式分解)
a(x1+x2)≥6
f(x1+x2)=1-2/[a(x1+x2)+1]≥1-2/(6+1)=5/7