如果知道Cauchy不等式,直接1/a+1/b+1/c = (a+b+c)(1/a+1/b+1/c) ≥ (1+1+1)² = 9.
如果只會(huì)均值不等式,就展開(kāi)1/a+1/b+1/c = (a+b+c)(1/a+1/b+1/c)
= 3+(a/b+b/a)+(b/c+c/b)+(c/a+a/c) ≥ 3+2+2+2 = 9.
已知a,b,c屬于R+,a+b+c=1,求證：1/a+1/b+1/c>=9
已知a,b,c屬于R+,a+b+c=1,求證：1/a+1/b+1/c>=9
數(shù)學(xué)人氣：912 ℃時(shí)間：2020-02-06 05:24:26
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