Xét hiệu \(\left(a_1+a_2+a_3\right)\left(b_1+b_2+b_3\right)-3\left(a_1b_1+a_2b_2+a_3b_3\right)\)
\(=a_1\left(b_1+b_2+b_3\right)+a_2\left(b_1+b_2+b_3\right)+a_3\left(b_1+b_2+b_3\right)-3a_1b_1-3a_2b_2-3a_3b_3\)
\(=a_1\left(b_1+b_2+b_3-3b_1\right)+a_2\left(b_1+b_2+b_3-3b_2\right)+a_3\left(b_1+b_2+b_3-3b_3\right)\)
\(=a_1\left(b_2+b_3-2b_1\right)+a_2\left(b_1+b_3-2b_2\right)+a_3\left(b_1+b_2-2b_3\right)\)
\(=a_1\left[\left(b_2-b_1\right)-\left(b_1-b_3\right)\right]+a_2\left[\left(b_3-b_2\right)-\left(b_2-b_1\right)\right]+a_3\left[\left(b_1-b_3\right)-\left(b_3-b_2\right)\right]\)
\(=a_1\left(b_2-b_1\right)-a_1\left(b_1-b_3\right)+a_2\left(b_3-b_2\right)-a_2\left(b_2-b_1\right)+a_3\left(b_1-b_3\right)-a_3\left(b_3-b_2\right)\)
\(=\left(a_1-a_2\right)\left(b_2-b_1\right)+\left(a_3-a_1\right)\left(b_1-b_3\right)+\left(a_2-a_3\right)\left(b_3-b_2\right)\)
Do giả thiết nên dễ thấy từng số hạng trên đều nhỏ hơn 0 nên tổng nhỏ hơn 0
=> ĐPCM
Dấu "=" khi \(\hept{\begin{cases}a_1=a_2=a_3\\b_1=b_2=b_3\end{cases}}\)
