AmerYR calculated $$f''(x)=\frac{a^2x^{a-1}\left(1+a+x^a(a-1)\right)}{(1-x^a)^3} - \frac{2}{(1-x)^3} \, .$$
As you noticed $f''(x)=0$ at $a=1$. So it suffices to show that $\partial_a f''(x) \leq 0$. In the comments I gave the derivative with respect to $a$ by $${\frac { \left( \left( \left( {a}^{2}-a \right) \ln \left( x \right) -3\,a+2 \right) {x}^{3\,a-1}+ \left( 4\,{a}^{2}\ln \left( x \right) -4 \right) {x}^{2\,a-1}+{x}^{a-1} \left( \left( {a}^{2}+a \right) \ln \left( x \right) +3\,a+2 \right) \right) a}{ \left( 1- {x}^{a} \right) ^{4}}} \, .$$ Since $\frac{(1-x^a)^4}{a \, x^{a-1}}>0$ we can multiply by this and obtain the objective $$\left( a\left( a-1 \right) {x}^{2\,a}+4\,{a}^{2}{x}^{a}+{a}^{2}+a \right) \ln \left( x \right) + \left( -3\,a+2 \right) {x}^{2\,a}-4 \,{x}^{a}+3\,a+2 \stackrel{!}{\leq} 0 \, .\tag{0}$$
We now need to get rid of the logarithm in order to get a manifest signature. I leave it to you to check that this expression goes to $-\infty$ as $x\rightarrow 0$, while it vanishes at $x=1$. Taking the derivative with respect to $x$ and dividing the resulting expression by $a \, x^{a-1}>0$ yields $$2\,a \left( \left( a-1 \right) {x}^{a}+2\,a \right) \ln \left( x \right) + \left( a+1 \right) {x}^{-a}+ \left( -5\,a+3 \right) {x}^{a} +4\,a-4 \stackrel{!}{\geq}0 \tag{1} \, .$$
Now rinse and repeat. Check the limiting cases (they are $+\infty$ and $0$), derive by $x$ and divide by $a\,x^{a-1}$ to obtain $$2a\left( a-1 \right) \ln \left( x \right) - \left( a+1 \right) {x}^{-2\,a}+4\,{x}^{-a}a-3\,a+1 \stackrel{!}{\leq} 0 \, .\tag{2}$$ Almost there; The limiting cases are manifestly $-\infty$ and $0$ respectively and so after deriving with respect to $x$ and multiplying by $\frac{x^{2a+1}}{2a}>0$ this becomes $$\left( a-1 \right) {x}^{2\,a}-2\,a\,{x}^{a}+a+1 \stackrel{!}{\geq} 0 \, .\tag{3}$$ One more time: At $x=0$ this is $a+1>0$ and at $x=1$ it vanishes again. Deriving with respect to $x$ and dividing by $2a\, x^{a-1}>0$ we have $$(a-1)x^a - a \stackrel{!}{\leq} 0 \, .\tag{4}$$
Best Answer
I am a newcomer so would prefer to just leave a comment, but alas I see no "comment" button, so I will leave my suggestion here in the answer box.
I have often used a Nelder-Mead "derivative free" algorithm (fminsearch in matlab) to minimize long and convoluted equations like this one. If you can substitute the constraint equation $g(\rho)$ into $f(\rho)$ somehow, then you can input this as the objective function into the algorithm and get the minimum, or at least a local minimum.
You could also try heuristic methods like simulated annealing or great deluge. Heuristic methods would give you a better chance of getting the global minimum if the solution space has multiple local minima. In spite of their scary name, heuristic methods are actually quite simple algorithms.
As for proving the concavity I don't see the problem. You mention in your other post that both $g(\rho)$ and $f(\rho)$ have first and second derivatives, so it should be straightforward, right?