用轮换对称式解决求最值得问题x>0,y>0.则(x/2x+y)+(y/x+2y)的最大值为?

用轮换对称式解决求最值得问题
x>0,y>0.则(x/2x+y)+(y/x+2y)的最大值为?

x/(2x+y)+y/(x+2y) = 1-(x+y)/(2x+y)+1-(x+y)/(x+2y)
= 2-(x+y)(1/(2x+y)+1/(x+2y))
= 2-1/3·((2x+y)+(x+2y))(1/(2x+y)+1/(x+2y))
≤ 2-1/3·(1+1)² (Cauchy不等式)
= 2/3.
可知x = y时等号成立,即最大值为2/3.