Question

If $x,y,z$ are positive then, minimum value of $x^{\log y-\log z}+y^{\log z-\log x}+z^{\log x-\log y}$ is

• A. 0
• B. 1
• C. 2
• D. 3

Answer 1

Answer: (D)

We know, that $AM\ge GM$
$\therefore$ $\frac{x^{\log y-\log z}+y^{\log z-\log x}+z^{\log x-\log y}}{3}\ge$
$\sqrt[3]{x^{\log y-\log z}.y^{\log z-\log x}.z^{\log x-\log y}}$ ..(i)
Now, $\log \{x^{\log y-\log z}.y^{\log z-\log x}.z^{\log x-\log y}\}$
$=\log \{x^{\log y}.x^{-\log z}.y^{\log z}.z^{-\log x}.z^{\log x}.z^{-\log y}\}$
$=\log (x^{\log y})+\log (x^{-\log z})+\log (y^{\log z})$
$+\log (y^{-\log x})+\log (z^{\log x})+\log (z^{-\log y})$
$=\log y\log x-\log z\log x+\log z\log y$
$-\log x\log y+\log x\log z-\log y\log z$
$=0$ $\Rightarrow$ $x^{\log y-\log z}+y^{\log z-\log x}+z^{\log x-\log y}\ge 3$
$\therefore$ FromEq. (i),
$x^{\log y-\log z}+y^{\log z-\log x}+z^{\log x-\log y}\ge 3$
Hence, minimum value is 3.