I am reading a paper in which they solve the following problem;
start
$$ \begin{aligned} \mathcal{W}_{\alpha_{1}, \beta_{1}, \theta} &=\frac{1}{2}\left[\cos \beta_{1} \cos \alpha_{1}+\cos \beta_{1} \cos \left(\theta-\alpha_{1}\right)+\cos \left(\beta-\beta_{1}\right) \cos \left(\frac{\theta}{2}-\alpha-\alpha_{1}\right)\right] \\ &+\frac{1}{2} \sqrt{4 \cos ^{2} \frac{\theta}{2}+1+4 \cos \frac{\theta}{2} \cos \beta \cos \alpha} \\ &+\sqrt{4 \cos ^{2} \frac{\theta}{2}+1-4 \cos \frac{\theta}{2} \cos \beta \cos \alpha} \\ &+\sqrt{4 \sin ^{2} \frac{\theta}{2}+1+4 \sin \frac{\theta}{2} \cos \beta \sin \alpha} \\ &+\sqrt{4 \sin ^{2} \frac{\theta}{2}+1-4 \sin \frac{\theta}{2} \cos \beta \sin \alpha} \end{aligned} $$ In order to obtain the maximal value of $\mathcal{W}$ expression only about the maximal guessing probability $\mathbf{p}=\frac{1+\cos \beta_{1} \cos \alpha_{1}}{2}$, we use the method of the extreme-value problem of multi-variable function and let $x=\cos \frac{\theta}{2}, y=\cos \beta, z=\cos \alpha$ and $u=\cos \alpha_{1}$ to apply to equation (shown above). Then we get
$$ \begin{aligned} \mathcal{W}_{\mathbf{p}}^{\max }=& \max _{\{(r, s, v, m)\}}\left\{(2 \mathbf{p}-1) r^{2}+(2 \mathbf{p}-1) \frac{\sqrt{1-m^{2}}}{m} r \sqrt{1-r^{2}}+\frac{(2 \mathbf{p}-1) s+\sqrt{1-s^{2}} \sqrt{m^{2}-(2 \mathbf{p}-1)^{2}}}{2 m}\right.\\ &\left[\left(r v+\sqrt{1-r^{2}} \sqrt{1-v^{2}}\right) m+\left(\sqrt{1-r^{2}} v-r \sqrt{1-v^{2}}\right) \sqrt{1-m^{2}}\right]+\frac{1}{2} \sqrt{4 r^{2}+1+4 r s v} \\ +&\left.\sqrt{4 r^{2}+1-4 r s v}+\sqrt{4\left(1-r^{2}\right)+1+4 \sqrt{1-r^{2}} s \sqrt{1-v^{2}}}+\sqrt{4\left(1-r^{2}\right)+1-4 \sqrt{1-r^{2}} s \sqrt{1-v^{2}}}\right\}, \end{aligned} $$ where $(r,s,v,m)$ is one of the real roots of equation set with variables $(x, y, z, u)$.
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How does one go about solving this? I have a similar problem to solve with a different $\mathcal{W}$ and $\mathbf{p}$. Is it always possible to find such a solution?