a·b=-1/2,c^2=(xa+yb)^2=x^2a^2+y^2b^2+2xyab=x^2+y^2+2xya·b=x^2+y^2-xy=1令x+y=t,y=t-x 代入x^2+y^2-xy=1式得：x^2+（t-x)2-x(t-x）=13x^2-3tx+t^2-1=0Δ=9t^2-12(t^2-1)≥0-3t^2+12≥0t^2≤4-2≤t≤2x+y的最大...