Assume $u > 1, p \in [0,1], \mu, \sigma$ are real numbers. Prove that the system has a unique solution $(p,u)$: $$ pu + \frac{1-p}{u} = \mu$$ $$ pu^2 + \frac{1-p}{u^2} = \sigma^2 + \mu^2$$
My attempt
I tried to square the first equation and subtract it from the second. The result was, after some cancellations, $$u^2 + \frac{1}{u^2} - 2p(1-p) = \sigma^2$$, which can be rewritten as a quadratic in $a = u^2$ : $$ a^2 - \frac{2p - 2p^2 + \sigma^2}{p(1-p)}a + 1 = 0$$ The determinant of this equation is always non-negative, so it always has two roots. Moreover, if we denote $b = \frac{2p - 2p^2 + \sigma^2}{p(1-p)}$, we can see that $2b \geq \sqrt{b^2 - 4}$ so the equation has both roots positive. This is where I am stuck. I am not ever sure if this is the right approach. Any help would be greatly appreciated.
Hint:
The system is equivalent to
$$\begin{cases}p(u^2-1)&=\mu u-1,\\p(u^2-1)(u^2+1)&=(\sigma^2+\mu^2)u^2-1.\end{cases}$$
Eliminating $p$, we have a cubic equation
$$\mu u^3-(\mu^2+\sigma^2+1)u^2-\mu u+2=0.$$
To prove that this equation has a single root $u>1$, you can evaluate the Sturm sequence of the polynomial at $u=1$ and $u=\infty$.