I am trying to understand the proof of Lemma 1 in this paper (Section 9.2).

The proof shows that given a discrete probability distribution $P=(p_1,p_2,...,p_k)$ where $p_1 \geq p_2 \geq ... \geq p_k$, and a discrete probability distribution $Q=(q_0,q_0,q_3,q_4...,q_k)$ where $q_0 = q_0 \geq q_3 \geq q_4 \geq ... \geq q_k$. Then for a fixed $P$, and a fixed $\alpha>1$, the minimizer of:

$$\min_{q_0,q_3,...,q_k} p_1(\frac{q_0}{p_1})^\alpha + p_2(\frac{q_0}{p_2})^\alpha + \sum_{i=3}^{k}p_i(\frac{q_i}{p_i})^\alpha$$

$$\text{such that:}\\ 2q_0 + q_3 + ... + q_k = 1\\q_i-q_0\leq 0 \quad i\geq 3\\-q_i\leq 0$$

Is given by $q_0 = (\frac{p_1^{1-\alpha}+p_2^{1-\alpha}}{2})^\frac{1}{1-\alpha}$ and $q_i = \frac{1-2q_0}{1-p_1-p_2}p_i$ for $i\geq 3$. I think I understand how these are found, however they then claim that plugging these quantities into the Renyi divergence formula:

$$D_\alpha := \frac{1}{1-\alpha}\log\bigg[p_1(\frac{q_0}{p_1})^\alpha + p_2(\frac{q_0}{p_2})^\alpha + \sum_{i=3}^{k}p_i(\frac{q_i}{p_i})^\alpha\bigg]$$

gives:

$$-\log\bigg[1-p_1-p_2 + 2(\frac{p_1^{1-\alpha} + p_2^{1-\alpha}}{2})^\frac{1}{1-\alpha}\bigg]\quad\quad\quad[*]$$

Perhaps I have misunderstood but I do not understand how $[*]$ is found. When I plug $q_0$ and $q_i$ into the Renyi divergence formula, I get:

\begin{equation*} \begin{split} D_\alpha &= \frac{1}{1-\alpha}\log\bigg[(p_1^{1-\alpha} + p_2^{1-\alpha})q_0^\alpha + \sum_{i=3}^{k}p_i(\frac{q_i}{p_i})^\alpha\bigg]\\ &= \frac{1}{1-\alpha}\log\bigg[(p_1^{1-\alpha} + p_2^{1-\alpha})(\frac{p_1^{1-\alpha} + p_2^{1-\alpha}}{2})^\frac{\alpha}{1-\alpha} + \sum_{i=3}^{k}p_i(\frac{q_i}{p_i})^\alpha\bigg]\\ &= \frac{1}{1-\alpha}\log\bigg[2(\frac{p_1^{1-\alpha} + p_2^{1-\alpha}}{2})^\frac{1}{1-\alpha} + \sum_{i=3}^{k}p_i(\frac{q_i}{p_i})^\alpha\bigg]\\ &= \frac{1}{1-\alpha}\log\bigg[2(\frac{p_1^{1-\alpha} + p_2^{1-\alpha}}{2})^\frac{1}{1-\alpha} + \sum_{i=3}^{k}p_i(\frac{1-2q_0}{1-p_1-p_2})^\alpha\bigg]\\ &= \frac{1}{1-\alpha}\log\bigg[2(\frac{p_1^{1-\alpha} + p_2^{1-\alpha}}{2})^\frac{1}{1-\alpha} + (1-p_1-p_2)^{1-\alpha}(1-2q_0)^\alpha\bigg]\\ &= \frac{1}{1-\alpha}\log\bigg[2(\frac{p_1^{1-\alpha} + p_2^{1-\alpha}}{2})^\frac{1}{1-\alpha} + (1-p_1-p_2)^{1-\alpha}(1-2(\frac{p_1^{1-\alpha}+p_2^{1-\alpha}}{2})^\frac{1}{1-\alpha})^\alpha\bigg]\quad\quad [!] \end{split} \end{equation*}

How do I get from this to $[*]$ or is there an obvious mistake I am making?

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Edit: Looking again at the KKT conditions, I'm not sure now how they arrive at $q_0 = (\frac{p_1^{1-\alpha}+p_2^{1-\alpha}}{2})^\frac{1}{1-\alpha}$ and $q_i = \frac{1-2q_0}{1-p_1-p_2}p_i$ for $i\geq 3$.

Let $Q = (q_0, q_0, q_3)$

Then the Lagrangian is \begin{equation*} \begin{split} L = p_1(\frac{q_0}{p_1})^\alpha + p_2(\frac{q_0}{p_2})^\alpha + p_3(\frac{q_3}{p_3})^\alpha + \mu_1(q_3-q_0) - \mu_2q_0 - \mu_3q_3 + \lambda(2q_0 + q_3 - 1) \end{split} \end{equation*}

Differentiating and setting the slack variables to 0, gives: \begin{equation*} \begin{split} & \lambda = -\alpha(\frac{q_3}{p_3})^{\alpha - 1}\\ & 2\lambda = -\alpha(\frac{q_0}{p_1})^{\alpha - 1} - \alpha(\frac{q_0}{p_2})^{\alpha - 1} \end{split} \end{equation*}

Then $q_0 = (\frac{p_1^{1-\alpha}+p_2^{1-\alpha}}{2})^\frac{1}{1-\alpha} \iff q_3 = p_3$. However, I don't understand why $q_3$ must equal $p_3$.