Đặt \(P=2ab+2bc+2abc-5ac\), ta sẽ chứng minh \(-15\le P\le7\)
Ta có:
\(P=2b\left(a+c\right)+2abc-5ac\le b^2+\left(a+c\right)^2+2abc-5ac\)
\(P\le a^2+b^2+c^2+2abc-3ac=6+2abc-3ac=ac\left(2b-3\right)+6\)
- Nếu \(b\le\dfrac{3}{2}\Rightarrow P< 6< 7\) (đúng)
- Nếu \(b>\dfrac{3}{2}\Rightarrow P\le\dfrac{1}{2}\left(a^2+c^2\right)\left(2b-3\right)+6=\dfrac{1}{2}\left(6-b^2\right)\left(2b-3\right)+6\)
\(\Rightarrow P\le7-\dfrac{1}{2}\left(b-2\right)^2\left(2b+5\right)\le7\)
Dấu "=" xảy ra khi \(\left(a;b;c\right)=\left(1;2;1\right)\)
Đồng thời:
\(P=2\left(ab+bc+abc\right)-5ac\ge-5ac\ge-\dfrac{5}{2}\left(a^2+c^2\right)=-\dfrac{5}{2}\left(6-b^2\right)=-15+\dfrac{5}{2}b^2\ge-15\)
Dấu "=" xảy ra khi \(\left(a;b;c\right)=\left(\sqrt{3};0;\sqrt{3}\right)\)
