\(P=ab+bc+ca-\frac{1}{2}abc=\frac{1}{3}\left(a+b+c\right)\left(ab+bc+ca\right)-\frac{1}{2}abc\)
\(P\ge\frac{1}{3}.9abc-\frac{1}{2}abc=\frac{5}{2}abc\ge0\)
\(P_{min}=0\) khi \(\left(a;b;c\right)=\left(0;0;3\right)\) và hoán vị
Áp dụng BĐT Schur: \(abc\ge\left(a+b-c\right)\left(a+c-b\right)\left(b+c-a\right)\)
\(\Leftrightarrow abc\ge\left(3-2a\right)\left(3-2b\right)\left(3-2c\right)\)
\(\Leftrightarrow abc\ge\left(3-2a\right)\left(9-6b-6c+4bc\right)\)
\(\Leftrightarrow abc\ge27-18\left(a+b+c\right)+12\left(ab+bc+ca\right)-8abc\)
\(\Leftrightarrow-3abc\le9-4\left(ab+bc+ca\right)\)
\(\Leftrightarrow-\frac{1}{2}abc\le\frac{3}{2}-\frac{2}{3}\left(ab+bc+ca\right)\)
\(\Rightarrow P\le ab+bc+ca+\frac{3}{2}-\frac{2}{3}\left(ab+bc+ca\right)\)
\(\Rightarrow P\le\frac{1}{3}\left(ab+bc+ca\right)+\frac{3}{2}\le\frac{1}{9}\left(a+b+c\right)^2+\frac{3}{2}=\frac{5}{2}\)
\(P_{max}=\frac{5}{2}\) khi \(a=b=c=1\)
