Originally Posted by

**SlipEternal** Cross-multiplying gives the following equations:

$\displaystyle (z-x)\log x = (y-z)\log y$ (Equation 1)

$\displaystyle (x-y)\log x = (y-z)\log z$ (Equation 2)

$\displaystyle (x-y)\log y = (z-x)\log z$ (Equation 3)

Adding the first two equations gives

$\displaystyle (z-y)\log x = (y-z)(\log y + \log z)$

Dividing both sides by $\displaystyle y-z$ gives

$\displaystyle -\log x = \log y + \log z$

So, $\displaystyle \log x + \log y + \log z = 0$ (Equation 4)

Next, from equation 3, adding $\displaystyle (y-x)\log y$ to both sides gives:

$\displaystyle (y-x)\log y + (z-x)\log z = 0$ (Equation 5)

Now, consider

$\displaystyle \begin{align*}x\log x + y\log y + z\log z & = x\log x + (y-x+x)\log y + (z-x+x)\log z \\ & = x(\log x + \log y + \log z) + (y-x)\log y + (z-x)\log z \\ & = (y-x)\log y + (z-x)\log z \text{ by Equation 4} \\ & = 0 \text{ by Equation 5}\end{align*}$

Hence, $\displaystyle x^x\cdot y^y\cdot z^z = 1$ as desired.