原式
=(z+a)*(x+a)/(x+y+a)-z*x/(x+y)-a
=(zx+ax+az+a^2)/(x+y+a)-zx/(x+y)-a
=(zx^2+ax^2+azx+a^2x+zxy+axy+azy+a^2y-zx^2-zxy-zxa)/((x+y)(x+y+a))-a
=(ax^2+a^2x+axy+azy+a^2y)/(x^2+2xy+y^2+ax+ay)-a
=(ax^2+a^2x+axy+azy+a^2y-ax^2-2axy-ay^2-a^2x-a^2y)/(x^2+2xy+y^2+ax+ay)
=(-axy+azy-ay^2)/(x^2+2xy+y^2+ax+ay)
=ay(z-x-y)/(x^2+2xy+y^2+ax+ay)
所以z>x+y,式子1大,如果z=x+y一样大,如果z