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This is my code

Solve[{1/2 (-(Sin[a]/Sqrt[3 + 2 (Cos[a] - Cos[b] - Cos[c])]) + Sin[a]/
      Sqrt[3 + 2 (-Cos[a] + Cos[b] - Cos[c])] + Sin[a]/Sqrt[
      3 + 2 (-Cos[a] - Cos[b] + Cos[c])] - Sin[a]/Sqrt[
      3 + 2 (Cos[a] + Cos[b] + Cos[c])]) == \[Lambda] (-(Sin[a]/(
      6 Sqrt[3 + 2 (Cos[a] + Cos[b] + Cos[c])]))), 
  1/2 (Sin[b]/Sqrt[3 + 2 (Cos[a] - Cos[b] - Cos[c])] - Sin[b]/Sqrt[
      3 + 2 (-Cos[a] + Cos[b] - Cos[c])] + Sin[b]/Sqrt[
      3 + 2 (-Cos[a] - Cos[b] + Cos[c])] - Sin[b]/Sqrt[
      3 + 2 (Cos[a] + Cos[b] + Cos[c])]) == \[Lambda] (-(Sin[b]/(
      6 Sqrt[3 + 2 (Cos[a] + Cos[b] + Cos[c])]))), 
  1/2 (Sin[c]/Sqrt[3 + 2 (Cos[a] - Cos[b] - Cos[c])] + Sin[c]/Sqrt[
      3 + 2 (-Cos[a] + Cos[b] - Cos[c])] - Sin[c]/Sqrt[
      3 + 2 (-Cos[a] - Cos[b] + Cos[c])] - Sin[c]/Sqrt[
      3 + 2 (Cos[a] + Cos[b] + Cos[c])]) == \[Lambda] (-(Sin[c]/(
      6 Sqrt[3 + 2 (Cos[a] + Cos[b] + Cos[c])]))), 
  1/2 (1 + Sqrt[3 + 2 (Cos[a] + Cos[b] + Cos[c])]/3) == k, 
  0 <= {a, b, c} <= Pi/2, Abs[a - b] <= c <= a + b,3 <= k <= 2 Sqrt[3]}, {a, b, 
  c, \[Lambda]}]

I obtained the equations from attempting to maximize a multivariable function with a constraint using Lagrange multiplier. The solution which satisfies the set of equations will be a critical point and I need to find a relation between a, b, c which maximizes my function. The relation that I should get is a=b=c (see this answer to my previous question), but I need to prove it analytically. But Mathematica can't solve the above set of equations. It does not show any error; it just keeps on running. I'm assuming there's something seriously wrong with my code or my method. Please help in any way possible.

EDIT

I have to maximize the function $$p(a,b,c)=\frac{1}{2} \left(\frac{1}{3} \sqrt{2 (\cos (a)+\cos (b)+\cos (c))+3}+1\right);$$ with the constarint that following function is a constant $k$ $$t(a,b,c)=\frac{1}{2} \left(\sqrt{2 (\cos (a)-\cos (b)-\cos (c))+3}+\sqrt{2 (-\cos (a)+\cos (b)-\cos (c))+3}+\sqrt{2 (-\cos (a)-\cos (b)+\cos (c))+3}+\sqrt{2 (\cos (a)+\cos (b)+\cos (c))+3}\right);$$ and $$0\leq\{a,b,c\}\leq\pi/2, |a-b|\leq c \leq a+b, 3\leq k \leq 2 \sqrt{3} $$

The code in the question is obtained when I attempted to solve it using the lagrange method.

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  • 1
    $\begingroup$ I doubt you can get analytical solution to such a complicated nonlinear set of equations. Maybe you need to look at numerical methods. $\endgroup$
    – Nasser
    Oct 20 '21 at 8:43
  • $\begingroup$ That's what I'm afraid of too. But I thought I could make some argument out of the obvious symmetry of the variables involved $\endgroup$
    – Dotman
    Oct 20 '21 at 8:51
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    $\begingroup$ Not sure if it helps much, but you can express $t$ entirely in terms of $p$ as follows: $t(a,b,c) = k = 3 (-2 + p(a, b, c) + p(a, b + π, c + π) + p(a + π, b, c + π) + p(a + π, b + π, c))$ $\endgroup$
    – flinty
    Oct 20 '21 at 13:21
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    $\begingroup$ ^ rearranging gives: $p(a,b,c) = (6 + k - 3 p(a, b + π, c + π) - 3 p(a + π, b, c + π) - 3 p(a + π, b + π, c))$. Since $k$ is constant, $p(a,b,c)$ is maximized when $p(a, b + π, c + π) + p(a + π, b, c + π) + p(a + π, b + π, c)$ is as small as possible (the -3p parts in the rearrangement must be small). Due to symmetry in the arguments (with the +Pi's), and symmetry of $p(a,b,c)$ cosine coefficients - the previous minimum will occur along the line a=b=c. Does that seem valid reasoning? $\endgroup$
    – flinty
    Oct 20 '21 at 13:42
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    $\begingroup$ @Dotman My intuition is you have four points in a tetrahedron formed by $(a,b,c),(a,b+\pi,c+\pi),(a+\pi,b,c+\pi),(a+\pi,b+\pi,c)$ with origin at $(a,b,c)$. Suppose to start with a1=b1=c1=0, then we can translate the tetrahedron along the line a=b=c sticking out from the origin into the middle of the face. If we deviate from this line at all, at least one of the 3 corner terms (the +Pi ones above) will dominate the sum which should be kept small (above) and lower $p(a,b,c)$. It's kind of hard to show outside of a hand-wavey argument, so you could try putting that on a more rigorous footing. $\endgroup$
    – flinty
    Oct 20 '21 at 18:03

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