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?

In the second solution, when $=$ in $\ge$ is satisfied, $a=b=c$. But then, $abc=17$ and $a+b=c$ makes the contradiction.
