An Inequality Problem with not nice conditions

As has been mentioned in some comments too, this question does not seem to allow a pre-calculus solution. Anyway, it is again a question of the symmetric type, such as listed in: Completed with: From the latter article comes the following
Theorem (The Purkiss Principle). Let $f$ and $g$ be symmetric functions with continuous second derivatives in the neighborhood of a point $P = (r, \cdots, r)$. On the set where $g$ equals $g(P)$, the function $f$ will have a local maximum or minimum at $P$ except in degenerate cases.
Our function $f$ in this case is: $$ f(a,b,c) = \frac{a^3}{a^2+b^2}+\frac{b^3}{b^2+c^2}+\frac{c^3}{c^2+a^2}$$ Applying the Purkiss Principle gives (with $r > 0$): $$ g(r,r,r)=r^2+r^2+r^2=3 \quad \Longrightarrow \quad r=1 \quad \Longrightarrow \quad f(r,r,r) = \frac{3}{2}$$ The main problem with the Purkiss Principle, most of the time, is to prove that the minimum found is global, making this part of the answer only a partial answer.

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.