证明函数f(x)=ax+b/x(a>0,b>0)在区间[根号b/a,+∞)上是增函数

x1,x2∈[√b/a,+∞),x1＜x2f(x1)-f(x2)=ax1+b/x1-ax2-b/x2=a(x1-x2)b（x2-x1)/x1x2=(x1-x2)(a-b/x1x2)=(x1-x2)(ax1x2-b)/x1`x2x1-x2＜0x1,x2∈[√b/a,+∞)x1x2＞b/aax1x2-b＞0(x1-x2)(ax1x2-b)/x1`x2＜0f(x1)-f(x2)...