$\newcommand\opt{\operatorname{opt}}\newcommand\de{\delta}\newcommand\si{\sigma}$The answer is no.

Indeed, take any $h\in(0,1)$. Let $X=\{1,2,3,4\}$. Suppose that $$\si_0=\frac12\de_1+\frac12\de_4,\quad\si_1=\frac12\de_2+\frac12\de_3$$ (where $\de_a$ is the Dirac probability measure supported on $\{a\}$), $$\rho(1,2)=\rho(3,4)=h,\quad\rho(1,3)=\rho(1,4)=\rho(2,3)=\rho(2,4)=1,$$ $$\rho'(1,3)=\rho'(2,4)=h,\quad\rho'(1,2)=\rho'(1,4)=\rho'(2,3)=\rho'(3,4)=1$$ (such metrics $\rho$ and $\rho'$ clearly exist). Then $\opt(\rho)\le h$ and $\opt(\rho')\le h$, whereas $(\rho+\rho')(i,j)\ge1+h$ for distinct $i$ and $j$ in $X$ and hence $\opt(\rho+\rho')\ge1+h$, so that $$\frac{\opt(\rho+\rho')}{\opt(\rho)+\opt(\rho')}\ge\frac{1+h}{2h}\to\infty$$ as $h\downarrow0$.

$\newcommand\opt{\operatorname{opt}}\newcommand\de{\delta}\newcommand\si{\sigma}$The answer is no.

Indeed, take any $h\in(0,1)$. Let $X=\{1,2,3,4\}$. Suppose that $$\si_0=\frac12\de_1+\frac12\de_4,\quad\si_1=\frac12\de_2+\frac12\de_3$$ (where $\de_a$ is the Dirac probability measure supported on $\{a\}$), $$\rho(1,2)=\rho(3,4)=h,\quad\rho(1,3)=\rho(1,4)=\rho(2,3)=\rho(2,4)=1,$$ $$\rho'(1,3)=\rho'(2,4)=h,\quad\rho'(1,2)=\rho'(1,4)=\rho'(2,3)=\rho'(3,4)=1$$ (such metrics $\rho$ and $\rho'$ clearly exist).

Then

• $\opt(\rho)\le h$ (witnessed by $\mu=\frac12\de_{(1,2)}+\frac12\de_{(4,3)}$)
• $\opt(\rho')\le h$ (witnessed by $\mu'=\frac12\de_{(1,3)}+\frac12\de_{(4,2)}$).

On the other hand, $(\rho+\rho')(i,j)\ge1+h$ for distinct $i$ and $j$ in $X$, whereas the support sets of $\si_0$ and $\si_1$ are disjoint. Therefore, $\opt(\rho+\rho')\ge1+h$, so that $$\frac{\opt(\rho+\rho')}{\opt(\rho)+\opt(\rho')}\ge\frac{1+h}{2h}\to\infty$$ as $h\downarrow0$.