Cho $a,b,c>0$. Chứng minh BĐT sau luôn đúng: $ab+bc+ac\ge (a+b+c)\sqrt[3]{abc}$
Cho $a,b,c>0$. Chứng minh BĐT $ab+bc+ac\ge (a+b+c)\sqrt[3]{abc}$
Started By hanguyen445, 25-09-2016 - 16:08
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