# 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:

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

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 !

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