L1:x-2y=0.L2:2x+y=0.===>d1=|a-2b|/√5,d2=|2a+b|/√5.d1+d2=4.===>|a-2b|+|2a+b|=4√5.由柯西不等式可知,（1+1)[(a-2b)²+(2a+b)²]≥(|a-2b|+|2a+b|)²=80.===>(a-2b)²+(2a+b)²≥40.===>5(a²+b²)≥40.===>a²+b²≥8.===>√(a²+b²)≥2√2.故[√(a²+b²)]min=2√2.