已知abc都是正数,求证:1/2a+1/2b+1/2c>=1/(a+b)+1/(a+c)+1/(b+c)
问题描述:
已知abc都是正数,求证:1/2a+1/2b+1/2c>=1/(a+b)+1/(a+c)+1/(b+c)
答
1/a+1/b≥2/√ab≥2/[(b+a)/2]=4/(b+a)(此处两个不等号均用了不等式x+y≥2√xy) 从而1/4a+1/4b≥1/(b+a) 同理1/4a+1/4c≥1/(c+a) 1/4b+1/4c≥1/(c+b) 相加得到1/2a+1/2b+1/2c≥1/(b+c)+1/(a+c)+1/(a+b)或者:1/4a+1/...