Finding the maximum of a function on $ \Bbb{S}^{7} $. I'm trying to find the maximum of the function
$$2 a^2 h+\sqrt{3} a d f+\sqrt{3} a e g+2 b^2 h-\sqrt{3} b d g+\sqrt{3} b e f\\+2 c^2
   h+\sqrt{3} c d^2+\sqrt{3} c e^2-\sqrt{3} c f^2-\sqrt{3} c g^2\\-d^2 h-e^2 h-f^2
   h-g^2 h-2 h^3$$
on the sphere $$a^2+b^2+c^2+d^2+e^2+f^2+g^2+h^2=1 \text{.}$$
I suspect it is $2$, but I can't come up with a proof. Lagrange multipliers don't work, since the resulting equations don't have an analytic solution.
The value $2$ is attained for $h=-1$.
My current idea is that terms of the form $x^2y$ can't be bigger than $\frac{2}{3\sqrt{3}}$ and terms of the form $xyz$ can't be bigger than $\frac{1}{3\sqrt{3}}$, but I'm not sure this helps.
I would also be satisfied with a slightly higher bound than $2$.
Additional information:
The function is proportional to $\sum_{ijk=1}^8 d_{ijk}p^i p^j p^k$ (the constraint is just $\sum_i (p^i)^2 = 1$), where $ \tau_i\tau_j = -\frac{2}{3}\delta_{ij} +(f_{ijk}-i\, d_{ijk})\tau_k$ holds for the generators $\tau_i$ of $SU(3)$ with $[\tau_i,\tau_j]=f_{ijk} \tau_k$ and the normalization condition $\mathrm{Tr}(\tau_i\tau_j) = -2\delta_{ij}$ is imposed.

(Update: I just realize that this is very similar to @Michael 's earlier solution.)
Your expression ($=:\Phi_0$) can be simplified using the following measures:
$$a=r\cos\alpha,\quad b=r\sin\alpha,\quad d=s\cos\beta,\quad e=s\sin\beta,\quad f=t\cos\gamma,\quad g=t\sin\gamma\ .$$
After some computation this results in
$$\Phi_1=2c^2h+\sqrt{3}c(s^2-t^2)-h(2h^2-2r^2+s^2+t^2)+\sqrt{3}rst\cos(\alpha-\beta+\gamma)\ .$$
Therefore we are left with the task of maximizing
$$\Phi_2:=2c^2h+\sqrt{3}c(s^2-t^2)-h(2h^2-2r^2+s^2+t^2)+\sqrt{3}rst$$
under the constraints
$$r\geq0, \quad s\geq0,\quad t\geq0,\quad r^2+s^2+t^2+c^2+h^2=1\ .$$
Using the main constraint we can rewrite $\Phi_2$ as
$$\Phi_3=-h^3+h\bigl(-1+3(c^2+r^2)\bigr)+\sqrt{3}\bigl(rst+c(s^2-t^2)\bigr)\ .$$
This suggests writing
$$c=x\cos\phi,\quad r=x\sin\phi\ ,$$
which then leads to
$$\Phi_4=-h(1+h^2-3x^2)+\sqrt{3}x\bigl((s^2-t^2) \cos\phi+st\sin\phi\bigr)\ ,\tag{1}$$
whereby the constraints now are
$$x\geq0, \quad s\geq0,\quad t\geq0,\quad x^2+s^2+t^2+h^2=1\ .$$
Maximizing $(1)$ over $\phi$ brings us to
$$\Phi_5=-h(1+h^2-3x^2)+\sqrt{3}x\sqrt{(s^2-t^2)^2+s^2 t^2}\ .$$
For given sum $s^2+t^2$ the square root on the right hand side is maximal when one of the variables is $=0$ (check this!), so that we arrive at
$$\Phi_6=-h(1+h^2-3x^2)+\sqrt{3}x s^2\tag{2}$$
under the constraints
$$x\geq0,\quad s\geq0,\quad x^2+s^2+h^2=1\ .$$
Eliminating $s^2$ from $(2)$ using the constraint leaves us with
$$\Phi_7=-h(1+h^2-3x^2)+\sqrt{3}x(1-h^2-x^2)\qquad(-1\leq h\leq 1, \ 0\leq x\leq\sqrt{1-h^2})\ .$$
This could be analyzed further, but a 3D plot of $\Phi_7$ as a function of $h$ and $x$ reveils that $\Phi_7$ indeed takes the maximal value $2$ when $h=-1$.
