Question

Cho ba số thực dương a, b, c thỏa mãn a+b+c=3. Chứng minh rằng

\(a^3+b^3+c^3+ab+ac+bc\ge6\)

Answer 1

\(a^3+a\ge2a^2\) ; \(b^3+b\ge2b^2\); \(c^3+c\ge2c^2\)

\(\Rightarrow a^3+b^3+c^3+3\ge2\left(a^2+b^2+c^2\right)\)

\(\Rightarrow a^3+b^3+c^3+ab+ac+bc\ge2\left(a^2+b^2+c^2\right)+ab+ac+bc-3\)

Mặt khác

\(P=2\left(a^2+b^2+c^2\right)+ab+ac+bc-3=\frac{3}{2}\left(a^2+b^2+c^2\right)+\frac{1}{2}\left(a+b+c\right)^2-3\)

\(P=\frac{3}{2}\left(a^2+b^2+c^2\right)+\frac{3}{2}\ge\frac{3}{2}.\frac{\left(a+b+c\right)^2}{3}+\frac{3}{2}=6\)

\(\Rightarrow a^3+b^3+c^3+ab+ac+bc\ge6\)

Dấu "=" xảy ra khi \(a=b=c=1\)