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))
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![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))](/uploads/image/z/12559966-70-6.jpg?t=%E5%B7%B2%E7%9F%A5a%2Cb%2Cc%E6%98%AF%E6%AD%A3%E6%95%B0%2C%E4%B8%94ab%2Bbc%2Bca%3D1%2C%E6%B1%82%E8%AF%81%EF%BC%9A%EF%BC%881%EF%BC%89a%2Bb%2Bc%EF%BC%9E%3D3%EF%BC%BE%281%2F2%29%282%29%E3%80%94a%2F%28bc%29%E3%80%95%EF%BC%BE%EF%BC%881%2F2%EF%BC%89%2B%5Bb%2F%28ac%29%5D%EF%BC%BE%281%2F2%29%2B%5Bc%2F%28ab%29%5D%EF%BC%BE%281%2F2%29%EF%BC%9E%3D3%EF%BC%BE%281%2F2%29%28a%EF%BC%BE%281%2F2%29%2Bb%EF%BC%BE%EF%BC%881%2F2%EF%BC%89%2Bc%EF%BC%BE%281%2F2%29%29)
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))
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))
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))
1
a²+b²≥2ab
b²+c²≥2bc
c²+a²≥2ca
两边一加得到
2(a²+b²+c²)≥2(ab+bc+ca)=2
所以a²+b²+c²≥1
a²+b²+c²+2ab+2ac+2ca=(a+b+c)²≥3
这样a+b+c≥3^(1/2)
2
逆证倒推
a/(bc)〕^(1/2)+[b/(ac)]^(1/2)+[c/(ab)]^(1/2)>=3^(1/2)(a^(1/2)+b^(1/2)+c^(1/2))
得到(a+b+c)/√abc≥3^(1/2)(a^(1/2)+b^(1/2)+c^(1/2))
这样因为a+b≥2√ab
b+c≥2√bc
a+c≥2√ac
所以a+b+c≥√ab+√bc+√ac
再结合第一问和结果OK!显然了
不等式难起来就恼火老~~~
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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))
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