Maximum of a multivariable function with constraints 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.
$$t(a,b,c)=\frac{1}{2} \left(\sqrt{2 (\cos (a)-\cos (b)-\cos (c))+3}+\sqrt{2 (-\cos (a)+\cos (b)-\cos (c))+3}+\sqrt{2 (-\cos (a)-\cos (b)+\cos (c))+3}+\sqrt{2 (\cos (a)+\cos (b)+\cos (c))+3}\right);$$
and
$$0\leq\{a,b,c\}\leq\pi/2,|a-b|\leq c \leq a+b $$
I believe the solution is when $a=b=c$ (see this answer), but I don't yet have a way to prove it. Is it possible to solve this using Lagrange multipliers? Kindly help any way you can.
EDIT
Can we use the symmetry in the objective and the constraint to prove this? As in, can we say that the critical points should satisfy certain symmetry and then proceed to show that the function is max at that point??
A related question on symmetry was answered here.
 A: Partial answer.
What @Domen did in the linked page looks quite convincing but we can make the result simpler. Let $x=\cos(a)$ which makes
$$p=p(a,a,a)=\frac{1}{2} \left(\frac{1}{3} \sqrt{6 x+3}+1\right)\tag1$$ $$t=t(a,a,a)=\frac{1}{2} \left(3 \sqrt{3-2 x}+\sqrt{6 x+3}\right)\tag2$$
From $(1)$
$$x=6 p^2-6 p+1$$ Plugging in $(2)$ leads to
$$t=\frac{3}{2} \left(\sqrt{-12 p^2+12 p+1}+\sqrt{(2 p-1)^2}\right)$$ This leads to a quartic equation in $p$
$$p^4-2 p^3+ \left(1+\frac{t^2}{36}\right)p^2-\frac{t^2}{36}p+\frac{t^2(t^2-9
  )}{1296}=0$$
the roots of which being explicit.
Selecting the proper one, as @Damen already had to do, the solution is
$$\color{blue}{p=\frac{1}{12} \left(t+6 +\sqrt{3(12-t^2)}\right)}$$ which is identical to
@Damen's expression (but a bit simpler).
Edit
If we consider Lagrange multipliers, that is to say
$$F=p(a,b,c)+\lambda(t(a,b,c)-k)$$ we have
$$\frac{\partial F}{\partial a}= Q_a\sin(a)\qquad\frac{\partial F}{\partial b}= Q_b\sin(b)\qquad\frac{\partial F}{\partial c}= Q_c\sin(c)$$ which are supposed to be zero. So $a=b=c=0$ is at least a trivial solution.
A: Remark: In the proof, we use the expression $\color{blue}{p=\frac{1}{12} \left(t+6 +\sqrt{3(12-t^2)}\right)}$ due to @Claude Leibovici which is simpler than @Domen's $p = \frac{1}{12}(\sqrt{2 t (\sqrt{36-3
   t^2}-t)+36}+6) $ in the linked page, though they are the same (easy to prove).

Consider the following optimization problem:
\begin{align*}
 &\max_{a, b, c} ~ p(a, b, c)\\
 &\mathrm{s.t.} \quad t(a, b, c) = T, \\
 &\qquad\ 0 \le a, b, c \le \pi/2, ~ |a - b| \le c \le a + b
\end{align*}
where $T$ is a constant such that the feasible region is not empty.
Denote its (global) maximum by $p(T)$.
As expected (to be proved), we have
$$p(T) = \frac{1}{12}\left(T + 6 + \sqrt{36 - 3T^2}\right).$$
(Note:
We have $t(a,b,c) = \frac12(x + y + z + w) \le \frac12\sqrt{4(x^2 + y^2 + z^2 + w^2)} = \sqrt{12}$ where $x, y, z, w$ denote $\sqrt{2 (\cos (a)-\cos (b)-\cos (c))+3}$ etc. Thus, $T \le \sqrt{12}$.)
Proof:
Fact 1: For all $0 \le a, b, c \le \pi/2, ~ |a - b| \le c \le a + b$,
$$p(a, b, c) \le \frac{1}{12}\left(t(a,b,c) + 6 + \sqrt{36 - 3[t(a,b,c)]^2}\right).$$
(The proof is given at the end.)
Using Fact 1, we have $p(T) \le \frac{1}{12}\left(T + 6 + \sqrt{36 - 3T^2}\right)$.
On the other hand, let
$$a = b = c = 
\arccos \left(1 - \frac{1}{12}T^2 + \frac{1}{12}\sqrt{36 - 3T^2}\right)$$
which satisfy $0 \le a, b, c \le \pi/2, ~ |a - b| \le c \le a + b$;  $~ t(a, b, c) = T$; and
$$p(a,b,c) = \frac{1}{12}\left(T + 6 + \sqrt{36 - 3T^2}\right).$$
As a result, we have
$$p(T) = \frac{1}{12}\left(T + 6 + \sqrt{36 - 3T^2}\right).$$
We are done.


Proof of Fact 1:
For convenience, use $p, t$ to denote $p(a,b,c), t(a,b,c)$.
It suffices to prove that
$$36 - 3t^2 \ge (12p - t - 6)^2$$
or
$$-144p^2 + 24pt - 4t^2 + 144p - 12t \ge 0. \tag{1}$$
Let
\begin{align*}
 x &= \sqrt{2 (\cos (a)-\cos (b)-\cos (c))+3}, \\
 y &= \sqrt{2 (-\cos (a)+\cos (b)-\cos (c))+3}, \\
 z &= \sqrt{2 (-\cos (a)-\cos (b)+\cos (c))+3}, \\
 w &= \sqrt{2 (\cos (a)+\cos (b)+\cos (c))+3}. \tag{2}
\end{align*}
We have
$$p = \frac12(w/3 + 1), \quad t = \frac12(x + y + z + w).$$
(1) becomes
$$36 - (x + y + z)^2 - 3w^2 \ge 0$$
which is true since
\begin{align}
 36 - (x + y + z)^2 - 3w^2 &\ge
 36 - 3(x^2 + y^2 + z^2) - 3w^2\\
 &= 36 - 3(x^2 + y^2 + z^2 + w^2)\\
 &= 0
\end{align}
where we have used $(x + y + z)^2 \le 3(x^2 + y^2 + z^2)$.
We are done.
