己知a,b,c∈R+,且a+b+c=1,证明:√ab+√bc+√ca≤1,证明:bc/a+c己知a,b,c∈R+,且a+b+c=1,证明:√ab+√bc+√ca≤1,证明:bc/a+ca/b+ab/c≥1
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己知a,b,c∈R+,且a+b+c=1,证明:√ab+√bc+√ca≤1,证明:bc/a+c己知a,b,c∈R+,且a+b+c=1,证明:√ab+√bc+√ca≤1,证明:bc/a+ca/b+ab/c≥1(图1)
======以下答案可供参考======
供参考答案1:
证明:因为(bc)^2+(ac)^2]+[(bc)^2+(ab)^2]+[(ac)^2+(ab)^2]>=2abc(a+b+c)=2ab
所以bc/a+ca/b+ab/c>=1