证明:设x1>x2≥0,则f(x1)-f(x2)=√(x1^2+1)-ax1-√(x^2+1)+ax2=(x1^2-x2^2)/[√(x1^12+1)+√(x2^2+1)]-a(x1-x2)=(x1-x2){x1+x2-a[√(x1^2+1)+√(x2^2+1)]}/[√(x1^2+1)+√(x2^2+1)]又x1>x2≥0,a≥1,即x1-x2>0,x1...=(x1^2-x2^2)/[√(x1^12+1)+√(x2^2+1)]-a(x1-x2)第二步看不懂~~~！！√(x1^2+1)-√(x^2+1)=(√(x1^2+1)-√(x^2+1))/1=(√(x1^2+1)-√(x^2+1))(√(x1^2+1)+√(x^2+1))/(√(x1^2+1)+√(x^2+1)).=(x1^2-x2^2)/[√(x1^12+1)+√(x2^2+1)]看懂了吗分子分母同乘(√(x1^2+1)+√(x^2+1))