Another possibility would have been to introduce spherical coordinates, thereby eliminating the equation $a^2+b^2+c^2=3$ :
$$
a = \sqrt{3}\cos(\theta)\cos(\phi) \qquad b = \sqrt{3}\cos(\theta)\sin(\phi) \qquad c = \sqrt{3}\sin(\theta)
$$
But this destroys the beautiful symmetry in the first place. And it doesn't help much in the second place, because
the partial derivatives to $\theta$ and $\phi$ are too messy to determine the values where they are zero, which
would be required for establishing a minimum (at least MAPLE - as steered by me - refuses to do the job).
EDIT. Which is a bit too pessimistic view; one can do something with those spherical coordinates:
$$
a = \sqrt{3}\cos(\theta)\cos(\phi)=1 \quad , \quad b = \sqrt{3}\cos(\theta)\sin(\phi)=1 \quad , \quad c = \sqrt{3}\sin(\theta)=1 \\
\Longrightarrow \quad \theta = \arcsin(1/\sqrt{3})=\arctan(1/\sqrt{2}) \quad \Longrightarrow \quad \phi = \arcsin(1/\sqrt{2})=\arctan(1)=\pi/4
$$
Substitution of the spherical coordinates in our function $f$ defines $\;g(\phi,\theta)=f(a,b,c)-3/2\;$.
After some tedious calculations (or rather trusting MAPLE :-) for $(\phi_0,\theta_0) = \arctan(1,1/\sqrt{2})$ :
$$
g(\phi_0,\theta_0) = 0 \quad ; \quad
\left[ \begin{array}{c} \frac{\partial g}{\partial \phi} \\ \frac{\partial g}{\partial \theta} \end{array} \right](\phi_0,\theta_0) = 0
\quad ; \quad
\left[ \begin{array}{cc} \frac{\partial^2 g}{\partial \phi^2} & \frac{\partial^2 g}{\partial \phi \, \partial \theta} \\
\frac{\partial^2 g}{\partial \theta \, \partial \phi} & \frac{\partial^2 g}{\partial \theta^2} \end{array} \right](\phi_0,\theta_0) =
\left[ \begin{array}{cc} 3 & 0 \\ 0 & 3 \end{array} \right]
$$
Proving that there is indeed a local extreme at $\;g(\phi_0,\theta_0)\;$ and that this extreme is a minimum.
The problem is, again, to prove that the minimum found is global, making this part of the answer a partial answer as well.