Question

For $x,y,z>0$ and $x>m,y>n,z>r,$ if $\left|\begin{array}{ccc}x& n& r\\ m& y& r\\ m& n& z\end{array}\right|=0,$ then the greatest value of $\frac{27xyz}{(x-m)(y-n)(z-r)}$ is

Answer 1

Given: $\left|\begin{array}{ccc}x& n& r\\ m& y& r\\ m& n& z\end{array}\right|=0$
Applying $R_1\to R_1-R_2$ and $R_2\to R_2-R_3$
$\left|\begin{array}{ccc}x-m& n-y& 0\\ 0& y-n& r-z\\ m& n& z\end{array}\right|=0$

$\Rightarrow (x-m)(y-n)z+n(x-m)(z-r)+m(n-y)(r-z)=0$
$\Rightarrow \frac{m}{x-m}+\frac{n}{y-n}+\frac{z}{z-r}=0$
$\Rightarrow \frac{x}{x-m}+\frac{y}{y-n}+\frac{z}{z-r}=2$

Let $\frac{x}{x-m},\frac{y}{y-n}$ and $\frac{z}{z-r}$ be three positive numbers.
Now, $A.M.\ge G.M.$
$\Rightarrow \frac{\frac{x}{x-m}+\frac{y}{y-n}+\frac{z}{z-r}}{3}\ge {\left(\frac{xyz}{(x-m)(y-n)(z-r)}\right)}^{1/3}$
$\Rightarrow \frac{xyz}{(x-m)(y-n)(z-r)}\le {\left(\frac{2}{3}\right)}^3$
$\Rightarrow \frac{27xyz}{(x-m)(y-n)(z-r)}\le 8$
Hence, the greatest value of $\frac{27xyz}{(x-m)(y-n)(z-r)}$ is $8$