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Let $u_1,u_2,\ldots ,u_n, t_1,t_2,\alpha, \beta \geq 0$ and consider a system of equations $$\begin{cases} \sum_{i=1}^n \left( \frac{u_i}\beta\right)^\alpha =t_1\\ \sum_{i=1}^n\left( \frac{u_i}{\beta}\right)^{-\alpha}=t_2\end{cases}.$$ I'm trying to prove\disprove that there is a unique solution $\alpha,\beta$. Equating the Jacobian determinant to $0$, got me $$\sum_{i=1}^n \ln\left(\frac{u_i}{\beta}\right)u_i^\alpha\sum_{i=1}^nu_i^{-\alpha}+\sum_{i=1}^nu_i^\alpha\sum_{i=1}^n\ln\left(\frac{u_i}{\beta}\right)u_i^{-\alpha}=0,$$ but there doesn't seem to be a way forward from there. Any alternative approaches to this problem?

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1 Answer 1

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Multiplying the equations we get $$\begin{cases} \sum_{i=1}^nu_i^\alpha \sum_{i=1}^n u_i^{-\alpha}=t_1t_2 \\ \sum_{i=1}^n\left( \frac{u_i}{\beta}\right)^{-\alpha}=t_2\end{cases}$$ and we know that the function $$p(\alpha)=\sum_{i=1}^nu_i^\alpha \sum_{i=1}^n u_i^{-\alpha}$$ is monotone, since sum and product of monotone functions retains the property. Monotonicity proves that the solution is unique.

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