If $c_1 \geq d_1, c_2 \geq d_2$ and $c_3 \geq d_3$ and we change the third coulumn of $ \det\begin{pmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\a_3 & b_3 & c_3 \end{pmatrix}$ as $ \det\begin{pmatrix} a_1 & b_1 & d_1 \\ a_2 & b_2 & d_2 \\a_3 & b_3 & d_3 \end{pmatrix}$, what will be the relation between these determinants. My intuition is: $$\det\begin{pmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\a_3 & b_3 & c_3 \end{pmatrix} \geq \det\begin{pmatrix} a_1 & b_1 & d_1 \\ a_2 & b_2 & d_2 \\a_3 & b_3 & d_3 \end{pmatrix}.$$
But I am not getting any reference for this.
Suppose this was true. Then $$det(a,b,c) \geq det(a,b,d)$$ $$-det(a,b,c) \leq -det(a,b,d)$$ $$det(-a,b,c) \leq det(-a,b,d)$$ provides a counterexample.