帮我解一道高中数学题，估计要用基础不等式，急

先完成这一步:
(x-y)^2>=0 ===> (x+y)^2-4xy>=0 ===> (x+y)^2>=4xy ===>
(x+y)/xy>=4/(x+y) ===> 1/y +1/x >= 4/(x+y) .
第二步:

1/2a+1/2b>=4/(2a+2b)=2/(a+b)
1/2c+1/2b>=4/(2c+2b)=2/(c+b)
1/2a+1/2c>=4/(2a+2c)=2/(a+c)
[1/2a+1/2b+1/2c+1/2b+1/2a+1/2c]/2 >= [2/(a+b)+2/(c+b)+2/(a+c)]/2
====>1/(2a)+1/（2b)+1/(2c)≥1/(b+c)+1/(c+a)+1/(a+b)