= Solution
For two <spin one-half> particles, $s_1^2=s_2^2=3\hbar^2/4$. Squaring $\mathbf S=\mathbf s_1+\mathbf s_2$ gives
$$
\boxed{f=a+\frac b2\left(S^2-\frac32\hbar^2\right).}
$$
Take $a,b$ real so that $f$ is <Hermitian>. Since $[S^2,S_z]=0$, the displayed <polynomial> implies $[f,S^2]=[f,S_z]=0$. There is a common <orthonormal> <eigenbasis>, so all three observables can be simultaneously certain \b[in a common <eigenstate>]. Commutation does not mean that every <quantum superposition> has definite values of all three.
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