<s>[INST] Given $0Z$ 
Given $x\in \mathbb R$, $0<x<1$, $\{m,n\}\subset \mathbb N$,$0<n<m$,
  $$Z=x^m-\frac{1}{x^m}~~\text{and}~~Y=x^n-\frac{1}{x^n}.$$
  Show that: (a) $Z$ and $Y$ are negative; (b) $Y>Z.$

Edit: the original exponents for $Y$ and $Z$ were switched in the original post. Now they are correct.
Problem from a book on inequalities at contest level. I got stuck on part (b), not able to find an argument for the general case.
My attempt: 
Part (a): developing $Z$ it follows that
$$Z=\frac{x^{2m}-1}{x^m}<0$$
as $0<x<1$, $x^{2m}-1<0$ and $x^m>0$. Similar argument holds for $Y$.
Part (b): only able to show for restricted cases, as $m=2$ and $n=1$, for instance:
$$Y=x-\frac{1}{x}~~\text{and}~~Z=x^2-\frac{1}{x^2}=(x-\frac{1}{x})(x+\frac{1}{x})=Y(x+\frac{1}{x})$$
as $Y<0$ and $x+\frac{1}{x}>1$, it follows that $Y>Z$, as required. I've got other restricted cases but I was not able generalize for all $0<n<m$. My intuition is suggesting an argument constructed on the identity
$$x^n-\frac{1}{x^n}=(x-\frac{1}{x})(x^{n-1}+x^{n-2}\frac{1}{x}+\ldots+\frac{1}{x^{n-1}})$$
but I have not find it so far.
Hints and solutions are appreciated. Sorry if this is a duplicate.
 [/INST] Since $0<n<m$ then we have $x^n>x^m$ because $0<x<1$.
Now suppose that we have $Y \geq Z$.
$Y \geq Z \implies x^m-\dfrac{1}{x^m} \geq x^n-\dfrac {1}{x^n} \implies \dfrac {x^{2m}-1}{x^m} \geq \dfrac {x^{2n}-1}{x^n} \implies \dfrac {1-x^{2m}}{x^m} \leq \dfrac{1-x^{2n}}{x^n} \implies x^n(1-x^{2m}) \leq x^m(1-x^{2n}) \implies x^n-x^{2m+n} \leq x^m - x^{2n+m} \implies x^n-x^m \leq x^{2m+n}-x^{2n+m}=x^{m+n}(x^m-x^n)$, 
which is a contradiction, since $x^n-x^m$ is positive and $x^{m+n}(x^m-x^n)$ is negative, so we have $Y<Z$.
</s>