数学题在线解答 设a,b,c均为正实数，求证1/2a+1/2b+1/2c大于等于1/(b+c)+1/(c+a)+1/(a+b)
展开全部
证明：∵1/x+1/y>-4/(x+y)=(x+y)/(xy)-4/(x+y)
=[(x+y)^2-4xy]/[(xy)(x+y)]
=(x-y)^2/[(xy)(x+y)]
当x>0,y>0时 (x-y)^2/[(xy)(x+y)]>=0
∴1/(4x)+1/(4y)>=1/(x+y)
从而 1/2a+1/2b+1/2c=1/4a+1/4b+1/4b+1/4c+1/4c+1/4a
>=1/(a+b)+1/(b+c)+1/(c+a)
∴1/2a+1/2b+1/2c大于等于1/(b+c)+1/(c+a)+1/(a+b)
=[(x+y)^2-4xy]/[(xy)(x+y)]
=(x-y)^2/[(xy)(x+y)]
当x>0,y>0时 (x-y)^2/[(xy)(x+y)]>=0
∴1/(4x)+1/(4y)>=1/(x+y)
从而 1/2a+1/2b+1/2c=1/4a+1/4b+1/4b+1/4c+1/4c+1/4a
>=1/(a+b)+1/(b+c)+1/(c+a)
∴1/2a+1/2b+1/2c大于等于1/(b+c)+1/(c+a)+1/(a+b)
本回答由网友推荐
已赞过
已踩过<
评论
收起
你对这个回答的评价是？
推荐律师服务：
若未解决您的问题，请您详细描述您的问题，通过百度律临进行免费专业咨询
