已知a,b,c是正数,且ab+bc+ca=1,求证:(1)a+b+c>=3^(1/2)
(2)〔a/(bc)〕^(1/2)+[b/(ac)]^(1/2)+[c/(ab)]^(1/2)>=3^(1/2)(a^(1/2)+b^(1/2)+c^(1/2))
人气:305 ℃ 时间:2021-03-04 02:46:47
解答
1a²+b²≥2abb²+c²≥2bcc²+a²≥2ca两边一加得到2(a²+b²+c²)≥2(ab+bc+ca)=2所以a²+b²+c²≥1a²+b²+c²+2ab+2ac+2ca=(a+b+c)²≥3这...
推荐
猜你喜欢