# How to prove the following optimization result from Mathematica?

Currently, I am working on a decision problem in Telecommunication. This problem is reformulated under a form of Optimization problem that is:

$$\begin{gathered} {\text{Minimize}}\,\,\,A \hfill \ st.\,\,\,\frac{{a\left( {\left( {{t^2} – 1} \right){S_1} + {{\left( {\frac{B}{A}} \right)}^2}{{\left( {{t^2} + 1} \right)}^2} – 4{t^2}} \right)}}{{\left( {{{\left( {\frac{B}{A}} \right)}^2} – 1} \right){{\left( {{t^2} + 1} \right)}^2}}} + D\left( {\frac{2}{{1 + {t^2}}} – 1} \right) \leqslant \frac{{2t}}{{{{\left( {{t^2} + 1} \right)}^2}}}\left( {a\left( {1 – {t^2}} \right) – {S_2}} \right) \hfill \ \end{gathered}$$

Where $$a > 0,A>0, B > 0, C > 0, 0 < \frac{B}{A} < 1$$ and the term $${S_1}$$ and $${S_2}$$ are given as:

$$\begin{gathered} {S_1}^2 = {\left( {\frac{B}{A}} \right)^2}{\left( {{t^2} + 1} \right)^2} – 4{t^2} \hfill \ {S_2}^2 = {\left( {\frac{1}{{BC}}} \right)^2}{\left( {{t^2} + 1} \right)^2} – 4{a^2}{t^2} \hfill \ \end{gathered}$$

Note that the term $$D$$ is guarantee to be negative ($$D<0$$). Also, the term $$t = \tan \left( {\frac{\theta }{2}} \right)$$ where the angle $$0 < \theta < \frac{\pi }{2}$$ which lead to $$0 < t < 1$$.

Practically speaking, there is also a constraint for how large and how small the value of $$A$$ should be that is $$0 < {M_L} < A < {M_U}$$. However, the physical range $$\left| {{M_U} – {M_L}} \right|$$ is quite large so this constraint can be dropped.

The followings is the code for invoking Mathematica parametric optimization:

Minimize[{A,(a*((t^2-1)Subscript[S, 1]+(B/A)^2 (t^2+1)^2-4t^2))/(((B/A)^2-1)*Power[t^2+1,2])+D*(2/(1+t^2)-1)<= (2*t)/Power[t^2+1,2] (a*(1-t^2)-Subscript[S, 2])&&D<0&&0<t<1 &&Power[Subscript[S, 2],2]==Power[1/(B*C),2]*(t^2+1)^2-4*a^2*t^2&&a>0&&B>0&&C>0&&A>0&&0<B/A<1&&Power[Subscript[S, 1],2]==(B/A)^2*(t^2+1)^2-4*t^2&&0<Subscript[M, L]<A<Subscript[M, U]&&Subscript[S, 1]>0&&Subscript[S, 2]>0},{A}] 

Which give the minimizer is $$A = \sqrt {\frac{{{B^2}{{\left( {1 + {t^2}} \right)}^2}}}{{4{t^2} + {{\left( {{S_1}} \right)}^2}}}}$$. An interesting fact is that Mathematica show that this minimizer can be achieve in the following cases:

Cases 1: $$t = – 1 + \sqrt 2$$ this means that $$\theta = \frac{\pi }{4}$$

Cases 2: $$0 < t < – 1 + \sqrt 2$$ so $$0 < \theta < \frac{\pi }{4}$$

Cases 3: $$– 1 + \sqrt 2 < t < 1$$ so $$\frac{\pi }{4} < \theta < \frac{\pi }{2}$$

Among all of these cases, this equality happens $$C = \sqrt {\frac{{1 + 2{t^2} + {t^4}}}{{{B^2}\left( {4{a^2}{t^2} + {S_2}^2} \right)}}}$$. I guess that this is an important result to construct the proof.

How to proof that this result is correct since Mathematica hide all the reasoning step ?

Thank you for your enthusiasm !