Question

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

• A. -1
• B. 8
• C. 1
• D. 2

Answer 1

The correct option is B 8

Given that,
$\left|\begin{array}{ccc}x& n& r\\ m& y& r\\ m& n& z\end{array}\right|=0\left(\begin{array}{c}{R}_1\to R_1-R_2\\ R_2\to R_2-R_3\end{array}\right)$
$\Rightarrow \left|\begin{array}{ccc}x-m& n-y& 0\\ 0& y-n& r-z\\ m& n& z\end{array}\right|=0$
$\Rightarrow (x-m)\left[z\right(y-n)-n(r-z\left)\right]+m\left[\right(n-y\left)\right(r-z\left)\right]=0$
$\Rightarrow \frac{z}{z-r}+\frac{n}{y-n}+\frac{m}{x-m}=0$
$\Rightarrow \frac{z}{z-r}+\frac{n}{y-n}+1+\frac{m}{x-m}+1=2$
$\Rightarrow \frac{z}{z-r}+\frac{n+y-n}{y-n}+\frac{m+x-m}{x-m}=2$
$\Rightarrow \frac{z}{z-r}+\frac{y}{y-n}+\frac{x}{x-m}=2$
Applying AM -GM Inequality
$\frac{\frac{x}{x-m}+\frac{y}{y-n}+\frac{z}{z-r}}{3}\ge {\left(\frac{xyz}{(x-m)(y-n)(z-r)}\right)}^{\frac{1}{3}}$
$\frac{xyz}{(x-m)(y-n)(z-r)}\le \frac{8}{27}$
So $\frac{27xyz}{(x-m)(y-n)(z-r)}\le 8$