设a>0,b>0,c>0,求证1/2a+1/2b+1/2c≥1/(b+c)+1/(c+a)+1/(a+b)

设a>0,b>0,c>0,求证1/2a+1/2b+1/2c≥1/(b+c)+1/(c+a)+1/(a+b)
设a>0,b>0,c>0,求证1/2a+1/2b+1/2c≥1/(b+c)+1/(c+a)+1/(a+b)

设a>0,b>0,c>0,求证1/2a+1/2b+1/2c≥1/(b+c)+1/(c+a)+1/(a+b)
1/4a+1/4b=(a+b)/4ab
而a>0,b>0,所以 (a+b)/2>=(ab)^0.5 即 (a+b)^2>=4ab => (a+b)/4ab>=1/(a+b),所以1/4a+1/4b>=1/(a+b)
同理
1/4a+1/4c>=1/(a+c)
1/4c+1/4b>=1/(c+b)
因此
1/4a+1/4b+1/4a+1/4c+1/4c+1/4b=1/2a+1/2b+1/2c>=1/(b+c)+1/(c+a)+1/(a+b)

1/4a+1/4b=(a+b)/4ab
而a>0,b>0，所以 (a+b)/2>=(ab)^0.5 即 (a+b)^2>=4ab => (a+b)/4ab>=1/(a+b)，所以1/4a+1/4b>=1/(a+b）
因此
1/4a+1/4b+1/4a+1/4c+1/4c+1/4b=1/2a+1/2b+1/2c>=1/(b+c)+1/(c+a)+1/(a+b)