 Solution

For two spin one-half particles, $s_1^2=s_2^2=3{\hslash }^2/4$. Squaring $\mathbf{S}={\mathbf{s}}_1+{\mathbf{s}}_2$ givesTake $a,b$ 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 in a common eigenstate. Commutation does not mean that every quantum superposition has definite values of all three.