Question

Cho a, b, c là 3 số không âm thỏa mãn a + b + c = 1. Chứng minh rằng: \(ab+bc+ca\le\dfrac{2}{7}+\dfrac{9abc}{7}\)

Answer 1

Đồng bậc : \(BDT\Leftrightarrow9abc+2\left(a+b+c\right)^3\ge7\left(ab+bc+ca\right)\left(a+b+c\right)\)

\(\Leftrightarrow2\left(a^3+b^3+c^3\right)\ge ab\left(a+b\right)+bc\left(b+c\right)+ca\left(c+a\right)\)

\(\Leftrightarrow\left(a+b\right)\left(a-b\right)^2+\left(b+c\right)\left(b-c\right)^2+\left(a+c\right)\left(c-a\right)^2\ge0\)( đúng)\(\Rightarrow DPcm\)

Dấu = xảy ra khi \(a=b=c=\dfrac{1}{3}\)