x>y>0，x+y≤2，∴2/(x+3y)+1/(x-y)=(√2)²/(x+3y)+1²/(x-y)≥(√2+1)²/[(x+3y)+(x-y)]=(3+2√2)/[2(x+y)]≥(3+2√2)/4.故(x+3y):√2=(x-y):1且x+y=2，即x=-1+2√2，y=3-2√2时，所求最小值为: (3+2√2)/4。
