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{{aleft( {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}}} + Dleft( {frac{2}{{1 + {t^2}}} – 1} right) leqslant frac{{2t}}{{{{left( {{t^2} + 1} right)}^2}}}left( {aleft( {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 !