Elementary algebra problem Consider the following problem (drawn from Stanford Math Competition 2014): "Find the minimum value of $\frac{1}{x-y}+\frac{1}{y-z}+ \frac{1}{x-z}$ for for reals $x > y > z$ given $(x − y)(y − z)(x − z) = 17.$"
Method 1 (official solution): Combining the first two terms, we have
$\frac{x−z}{(x-y)(y-z)} + \frac{1}{x-z}= \frac{(x-z)^2}{17}+ \frac{1}{x-z}.$
What remains is to find the minimum value of $f(a) = \frac{a^2}{17} + \frac{1}{a} = \frac{a^2}{17} + \frac{1}{2a}+ \frac{1}{2a}$ for positive values of $a.$ Using AM-GM, we get $f(a) \geq \frac{3}{68^{1/3}}$.
Method 2: Let $x-y:=a, \; y-z:=b, \; x-z:=c.$  Then $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}= \frac{ab+bc+ac}{17}= \frac{3}{17}\frac{ab+bc+ac}{3} \geq \frac{3}{17} (a^2b^2c^2)^{1/3} =\frac{3}{17^{1/3}}.$
So Method 2 seems to give a sharper bound than the official solution.  Have I done something wrong?

The official solution is wrong. The equality happens when $$\frac{a^2}{17}=\frac1{2a}$$ or $a^3=\frac{17}2<a^3$. Note that in your solution, the equality does not happen either.
Edit: I would solve the problem as follows.
Let $a=x-y,b=y-z$ then $ab(a+b)=17$ and we need to minimize
$$\frac1a+\frac1b+\frac1{a+b}=\frac{(a+b)^2+ab}{ab(a+b)}=\frac{(a+b)^2+ab}{17}.$$
We then need to minimize $$f(t) = (a+b)^2+ab=t^2+\frac{17}{t},$$ where $a+b=t$ satisfying $$t^2\ge 4ab=\frac{68}{t},\text{ or } t\ge \sqrt[3]{68}.$$
It's easy to see that
$$f'(t)=2t-\frac{17}{t^2}=\frac{2t^3-17}{t^2}>0.$$
So $f(t)$ attains its minimum at $t=\sqrt[3]{68}$. That is $x-z=\sqrt[3]{68}$ and $x-y=y-z=\frac{\sqrt[3]{68}}{2}$ and we see that the minimum is $\displaystyle\color{red}{\frac{5}{\sqrt[3]{68}}}$.
