Question

$\text{If}a,b,c,d\in R^{+}\text{and}a+b+c+d=2,\text{then the maximum value of}\left|\begin{array}{ccc}a-c^2& ab& a^2-a{c}^2-c\\ 1+c& 1+b& 1+a+ac\\ b+bc& b^2-d& ab+b+abc\end{array}\right|\text{is}$

Answer 1

$\Delta =\left|\begin{array}{ccc}a-c^2& ab& a^2-a{c}^2-c\\ 1+c& 1+b& 1+a+ac\\ b+bc& b^2-d& ab+b+abc\end{array}\right|C_3\to C_3-a{C}_1=\left|\begin{array}{ccc}a-c^2& ab& -c\\ 1+c& 1+b& 1\\ b+bc& b^2-d& b\end{array}\right|C_1\to C_1-c{C}_3=\left|\begin{array}{ccc}a& ab& -c\\ 1& 1+b& 1\\ b& b^2-d& b\end{array}\right|C_2\to C_2-b{C}_1=\left|\begin{array}{ccc}a& 0& -c\\ 1& 1& 1\\ b& -d& b\end{array}\right|=a(b+d)-c(-d-b)\Delta =ab+ad+bc+cd\text{Using AM-GM inequality, we have}\frac{(a+c)+(b+d)}{2}\ge \sqrt{(a+c)(b+d)}\Rightarrow \frac{2}{2}\ge \sqrt{ab+ad+bc+cd}\Rightarrow 1\ge ab+ad+bc+cd\therefore \text{Max. value of}\Delta =1$