已知a,b,c∈R,且ab+bc+ca=1,用综合法证明下列不等式成立的是

2024-11-27 23:41:52
推荐回答(1个)
回答1:

ab+bc+ca=1即2ab+2bc+2ca=2<=a²+b²+b²+c²+a²+c²=
2(a²++b²+c²),
∴a²++b²+c²>=1,A错
将2ab+2bc+2ca=2与a²++b²+c²>=1左左相加,右右相加,得(a+b+c)²≥3,B对,D错是因为
a+b+c可能为负,即a+b+c<=-√3
因为D已证出√(a+b+c)≥√3或a+b+c<=-√3
所以C.1/a+1/b+1/c≥2√3肯定是错的了

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