-Ta c/m BĐT: \(a^3+b^3+1\ge3ab\)
\(\Leftrightarrow\left(a+b\right)^3+1-3ab\left(a+b\right)-3ab\ge0\)
\(\Leftrightarrow\left(a+b+1\right)\left[\left(a+b\right)^2-a-b+1\right]-3ab\left(a+b+1\right)\ge0\)
\(\Leftrightarrow\left(a+b+1\right)\left(a^2+2ab+b^2-a-b+1-3ab\right)\ge0\)
\(\Leftrightarrow\left(a+b+1\right)\left(a^2+b^2-ab-a-b+1\right)\ge0\)
\(\Leftrightarrow\left(a+b+1\right)\left(2a^2+2b^2-2ab-2a-2b+2\right)\ge0\)
\(\Leftrightarrow\left(a+b+1\right)\left[\left(a-b\right)^2+\left(a-1\right)^2+\left(b-1\right)^2\right]\ge0\) (đúng)
-Quay lại bài toán:
\(\left(a^3+b^3+1\right)+\left(b^3+c^3+1\right)+\left(c^3+a^3+1\right)\ge3ab+3bc+3ca\)
\(\Leftrightarrow2\left(a^3+b^3+c^3\right)+3\ge3\left(ab+bc+ca\right)=3.3=9\)
\(\Leftrightarrow a^3+b^3+c^3\ge3\left(đpcm\right)\)
