Generalization of Minkowski inequality I am wondering if the following is true:

Suppose continuous function $g: [0, \infty) \to [0, \infty)$ satisfying $g(0)=0$ is increasing and strictly convex and (therefore) invertible. Let $||f ||^g$ denote $g^{-1}(\int_X g \circ |f| d \mu)$ where $f$ is measurable for measurable space $(X, M, \mu)$. Then for any measurable functions $f_1$ and $f_2$ we have $$ || f_1 +f_2||^g \le ||f_1||^g + ||f_2||^g$$

I think this is true, but am particularly interested if it is true for when $g(x) = (x+1)^p -1$ with $(X, M, \mu)$ being the usual Lebesgue measure space with $X=R^n$.
Any insights, ideas, or references would be appreciated.
 A: It's not true as written.  Counterexample: Let $X=[0,1]$ with Lebesgue measure.  Let $f_1(x)=f_2(x)=x$.  Let $g$ be a convex function whose graph goes through $(0,0)$, $(\frac32,\varepsilon)$, $(2,1)$, and is linear in between those points.  (That's not strictly convex, but we can perturb the example a bit to get strict convexity if we insist.)  Then
\begin{align*}
\|f_1\|_g+\|f_2\|_g &= 2g^{-1}\left(\int_0^1 g(t)\,dt\right) = 1 \\
\text{but}\quad
\|f_1+f_2\|_g &= g^{-1}\left(\frac12\int_0^2 g(t)\,dt\right) > \frac32
\end{align*}
(because the average value of $g$ on $[0,2]$ is greater than $\epsilon$).

But the thing is, since $\|\cdot\|_g$ isn't necessarily homogeneous, it's a bit unnatural to ask for the triangle inequality in the form $\|f_1+f_2\|\le\|f_1\|+\|f_2\|$.  I think we should instead try for
$$ \|(1-\lambda)f_1+\lambda f_2\|\le (1-\lambda)\|f_1\|+\lambda\|f_2\| $$
(for all $\lambda\in[0,1]$).

Update: Alas, that's not true either.  Counterexample: Let $X=[0,2]$ with Lebesgue measure.  Let $f_1=0$ and $f_2=1$.  Let $\lambda=\frac12$.  Let
$$ g(x) = \begin{cases}
  x &\text{if $0\le x\le 1$,} \\
  1+\frac1\varepsilon(x-1) &\text{if $x\ge 1$.}
\end{cases} $$
Then
\begin{align*}
(1-\lambda)\|f_1\|_g + \lambda\|f_2\|_g &= \tfrac12 g^{-1}(2g(1)) = \tfrac12(1+\varepsilon)
\\
\text{but}\quad
\|(1-\lambda)f_1+\lambda f_2\|_g &= g^{-1}(2g(\tfrac12)) = 1
\end{align*}
