Do \(0\le a;b;c\le1\Rightarrow ab\ge abc\Rightarrow\frac{ab+bc+ca-abc}{a+2b+c}\ge\frac{bc+ca}{a+2b+c}\ge0\)
\(P_{min}=0\) khi \(\left(a;b;c\right)=\left(0;0;1\right)\) và hoán vị
Tìm max:
\(P=\frac{ab+bc+ca-abc}{a+2b+c}=\frac{\left(a+b+c\right)\left(ab+bc+ca\right)-abc}{a+2b+c}=\frac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{a+2b+c}\)
\(P\le\frac{1}{4}.\frac{\left(a+2b+c\right)^2\left(c+a\right)}{\left(a+2b+c\right)}=\frac{\left(a+2b+c\right)\left(a+c\right)}{4}=\frac{\left(1+b\right)\left(1-b\right)}{4}\)
\(P\le\frac{1-b^2}{4}\le\frac{1}{4}\)
Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}a=c=\frac{1}{2}\\b=0\end{matrix}\right.\)
