Find minimum of $ [ x^{(\ln y-\ln z)} + y^{(\ln z-\ln x )} + z^{(\ln x-\ln y)} ]$

Question

If $$x\gt0,y\gt 0, z\gt 0$$ then find the minimum value of
$$ \left[ x^{(\ln y-\ln z)} + y^{(\ln z-\ln x )} + z^{(\ln x-\ln y)} \right]$$ Today I came across this question ( maybe it's from Progressions and Series ) and was unable to solve it , I tried in this way... $$ a^{log_cb}=b^{log_ca}$$hence $$ \left[ x^{(\ln y-\ln z)} + y^{(\ln z-\ln x )} + z^{(\ln x-\ln y)} \right]=\left[ \left({y\over z}\right)^{\ln x} +\left({z\over x}\right)^{\ln y} +\left({x\over y}\right)^{\ln z} \right] $$ now using property $AM \ge GM$ implies $$\left[ x^{(\ln y-\ln z)} + y^{(\ln z-\ln x )} + z^{(\ln x-\ln y)} \right]\ge \sqrt[3]{y^{\ln {x\over z}}z^{\ln {y\over x}}x^{\ln {z\over y}}} $$ but i cann't get any further .Is the question wrong? Is the approach wrong? Looking for some hints to get started...

Answer 1

The given expression can be rewritten as

$$\frac{x^{ln\left(y\right)}}{x^{ln\left(z\right)}}+\frac{y^{ln\left(z\right)}}{y^{ln\left(x\right)}}+\frac{z^{ln\left(x\right)}}{z^{ln\left(y\right)}}$$

Apply A.M >= G.M.

$$\left(\frac{\frac{x^{ln\left(y\right)}}{x^{ln\left(z\right)}}+\frac{y^{ln\left(z\right)}}{y^{ln\left(x\right)}}+\frac{z^{ln\left(x\right)}}{z^{ln\left(y\right)}}}{3}\right)\ge \left(\frac{x^{ln\left(y\right)}}{x^{ln\left(z\right)}}\cdot \frac{y^{ln\left(z\right)}}{y^{ln\left(x\right)}}\cdot \frac{z^{ln\left(x\right)}}{z^{ln\left(y\right)}}\right)^{\frac{1}{3}}$$

Using $a^{\log _c\left(b\right)}=b^{log_c\left(a\right)}$ , the denominators on the right side of the inequality can be manipulated.

$$\left(\frac{\frac{x^{ln\left(y\right)}}{x^{ln\left(z\right)}}+\frac{y^{ln\left(z\right)}}{y^{ln\left(x\right)}}+\frac{z^{ln\left(x\right)}}{z^{ln\left(y\right)}}}{3}\right)\ge \:\left(\frac{x^{ln\left(y\right)}}{z^{ln\left(x\right)}}\cdot \:\frac{y^{ln\left(z\right)}}{x^{ln\left(y\right)}}\cdot \:\frac{z^{ln\left(x\right)}}{y^{ln\left(z\right)}}\right)^{\frac{1}{3}}$$

$$\left(\frac{\frac{x^{ln\left(y\right)}}{x^{ln\left(z\right)}}+\frac{y^{ln\left(z\right)}}{y^{ln\left(x\right)}}+\frac{z^{ln\left(x\right)}}{z^{ln\left(y\right)}}}{3}\right)\ge \:\left(1\right)^{\frac{1}{3}}$$

$$\left(\frac{x^{ln\left(y\right)}}{x^{ln\left(z\right)}}+\frac{y^{ln\left(z\right)}}{y^{ln\left(x\right)}}+\frac{z^{ln\left(x\right)}}{z^{ln\left(y\right)}}\right)\ge \:\left(3\right)$$

Therefore the minimum value of the expression is 3.

Answer 2

Let

$$x=e^u \quad y=e^v \quad z=e^w \quad u,v,w\in \mathbb{R}$$

then

$$ x^{(\ln y-\ln z)} + y^{(\ln z-\ln x )} + z^{(\ln x-\ln y)} = \frac{e^{uv}}{e^{uw}}+\frac{e^{vw}}{e^{uv}}+\frac{e^{uw}}{e^{vw}}\ge 3$$

by AM-GM or more directly by Rearrangement inequality.