I believe this is the answer but I have trouble convincing myself that this is true and need help proving so.
X and Y are random variables with the same variance. Let Z be a random variable that is X with probability 1/2 and Y with probability 1/2. Also let
The unique $argmax_q E[S(q,X)] = Var(X)$
The unique $argmax_q E[S(q,Y)] = Var(Y)$
Then I think that $argmax_q E[S(q,Z)] = Var(X) = Var(Y)$, but am lacking a formal explanation of why.
I wanted to say that $argmax_q E[S(q,Z)] = argmax_q (1/2E[S(q,X)] + 1/2E[S(q,Y)])$ but am not sure what formal reasoning results in this conclusion.
You have $Z=BX+(1-B)Y$ with $B\sim \mathrm{Bernoulli}(1/2)$ and $X,Y,B$ mutually independent.
Fix any $q$. $$ \mathbb{E}[S(q,Z)] = \mathbb{E}[\mathbb{E}[S(q,BX+(1-B)Y)|B]] $$ Now, for any fixed $b\in\{0,1\}$, $$ \mathbb{E}[S(q,BX+(1-B)Y)|B] = \begin{cases} S(q,X) &\text{ if } b=1\\ S(q,Y) &\text{ if } b=0 \end{cases} = BS(q,X)+(1-B)S(q,Y) $$ and so $$\begin{align} \mathbb{E}[S(q,Z)] &= \mathbb{E}[B \mathbb{E}[S(q,X)]+(1-B)\mathbb{E}[S(q,Y)]] = \mathbb{E}[B]\mathbb{E}[S(q,X)]+\mathbb{E}[(1-B)]\mathbb{E}[S(q,Y)]\\ &= \frac{1}{2}\mathbb{E}[S(q,X)] + \frac{1}{2}\mathbb{E}[S(q,Y)] \end{align}$$ where we used independence of $X,B$ and that of $Y,B$.
So, that's a long-winded way of proving the inequality you saw was intuitively true.: $$ \forall q, \qquad \mathbb{E}[S(q,Z)] =\frac{1}{2}\mathbb{E}[S(q,X)] + \frac{1}{2}\mathbb{E}[S(q,Y)] \tag{1} $$