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因为a+2b+3c=1,所以
ab/3+bc+ca/2=(ab/3+bc+ca/2)*(a+2b+3c)
=a2b/3+2ab2/3+abc+a2c/2+abc+3ac2/2+abc+2b2c+3bc2
=(a2b/3+3bc2)+(2ab2/3+3ac2/2)+(a2c/2+2b2c)+3abc
括号内用均值不等式,即得
(a2b/3+3bc2)+(2ab2/3+3ac2/2)+(a2c/2+2b2c)大于等于6abc
代回,得证
1年前
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