No. Denote $f(x)=a$, $f(y)=b$, then $g(f^{-1}(t))=h(t)$. Applying $g$ to both parts of first inequality we get $h(\frac13 a+\frac23 b)&lt;\frac13 h(a)+\frac23h(b)$. It means that $h$ is convex (proof below), analogously the second inequality means that $h$ is concave.

Lemma. If $h$ is continuous on $[0,1]$ and $h(tx+(1-t)y)&lt;th(x)+(1-t)h(y)$ for fixed $t\in (0,1)$ and all $x&lt;y$, then $h$ is convex.

Proof. Consider the set $S$ of $s\in [0,1]$ such that $h(sx+(1-s)y)\leq s\cdot h(x)+(1-s)\cdot h(y)$ for $x\leq y$. It is clearly closed since $h$ is continuous. Assume that $S\ne [0,1]$, then the complement $[0,1]\setminus S$ is open, take any (maximal by inclusion) interval $(\alpha,\beta)\subset [0,1]\setminus S$, $\alpha\in S,\beta\in S$. We easily get $t\alpha+(1-t)\beta\in S$, a contradiction.

Symmetric inequality on generalized means
Is the intersection of two star-shaped sets star-shaped?

I don't think so but can't think of an example.
