Showing that $x_{11}x_{22}-x_{12}x_{21}>0$ given $x_{11}>x_{21}$ and $x_{12}>x_{22}$.

Question

Let $x_{11},x_{12},x_{21},x_{22}$ be positive numbers between zero and one, and $x_{11}>x_{21}$ and $x_{12}>x_{22}$. Also, $x_{11}+x_{21}=1$ and $x_{22}+x_{12}=1$.

I want to show that $x_{11}x_{22}-x_{12}x_{21}>0$.

I do the following:

$$x_{11}>x_{21}\Leftrightarrow x_{11}x_{22}x_{12}>x_{21}x_{22}x_{12},$$ which is true since $x_{22}$ and $x_{12}$ are positive.

Next I define $A=x_{11}x_{22}$ and $B=x_{12}x_{21}$. Thus from the previous step, $x_{12}A>x_{22}B$. I know that $x_{12}>x_{22}$, can I infer that because of that, $A>B$, and therefore conclude that my inequality is indeed positive?

Is what I am doing correct? If yes, is there any more "elegant" way to show it?

Thanks in advance.

edit: sorry for the confusing notation. I forgot to state that $x_{11}+x_{21}=1$ and $x_{22}+x_{12}=1$

Answer 1

It is not true. Take $x_{11}=\frac34$, $x_{12}=\frac45$, $x_{21}=\frac14$, and $x_{22}=\frac15$. Then $x_{11}x_{22}-x_{12}x_{21}=\frac34-\frac45<0$.