Minimizing $\sum\limits_{k=1}^{n} \; (x_k +1)^{{A/x_k}}$. Consider $ 3 $ positive real numbers such that $ abc = 2 $. 
(a) Show that $ f(a,b,c) = \sqrt[a]{a^2+2a+1} + \sqrt[b]{b^2+2b+1} + \sqrt[c]{c^2+2c+1} \geq 9. $
(b) Is there a triple $ (a,b,c) $ satisfying the equality case? If yes, find it. If not, find the minimum of $ f(a,b,c)  $.
Bernoulli's has worked just fine on (a), since $ \require{cancel} (1+a)^{\cancelto{2/a}{bc}} \geq 1 + \cancelto{2}{abc} \Leftrightarrow \sqrt[a]{a^2 + 2a +1} \geq 3 $. However, I could not find the equality case (by guessing-setting $ a=b=c$, of course) or a better minimum of $ f(a,b,c) $.
Moreover, given $ n $ positive real numbers such that $ \displaystyle \prod_{k=1}^{n} x_k  = P > 1, $ can we find the exact minimum of $  \displaystyle {g(n,P) = \sum_{k=1}^{n} \; (x_k +1)^{\displaystyle {P/x_k}}} $ and the $n$-tuple that yields that result?
PS: The generic-case answer is more preferable that the $ g(3,2) $ one.

We need to prove that $\sum\limits_{cyc}(1+a)^{\frac{2}{a}}\geq9$.
Indeed, let $a=\sqrt[3]2e^x$, $b=\sqrt[3]2e^y$ and $c=\sqrt[3]2e^z$.
Hence, $x+y+z=0$ and we need to prove that $\sum\limits_{cyc}f(x)\geq0$, where $f(x)=\left(1+\sqrt[3]2e^x\right)^{\frac{\sqrt[3]4}{e^x}}.$
But $f''(x)>0$ for all $x\geq0$, which says that by Vasc's RCF Theorem 
it remains to prove our inequality for $y=x$ and $z=-2x$ or
$$2(a+1)^{\frac{2}{a}}+(1+2/a^2)^{a^2}\geq9,$$
which is true and even 
$$2(a+1)^{\frac{2}{a}}+(1+2/a^2)^{a^2}\geq2+e^2$$
is true.
Done!
