│a+bt│^2=│a│^2+t^2│b│^2+2abt
根据二次函数性质:当t=-ab/│b│^2时,│a+bt│^2有最小值
b(a+tb)=ab+tb^2=ab-ab/│b│^2*b^2=ab-ab=0
此时b垂直于(a+tb)
故得证