Let $p,q,r:p+q+r=1$ be probabilities of faces $0,1,2$ mod 3 as in the above solution. The three rolls are either congruent mod 3, with probability $p^3+q^3+r^3$ or equivalent to 0, 1, 2 mod 3 in some order, with probability $\sum_{\mathrm{sym}}pqr=6pqr$. Then \[ p^3+q^3+r^3+6pqr \geq \frac14 \] \[ \Leftrightarrow 4(p^3+q^3+r^3+6pqr) \geq (p+q+r)^3 \] \[ \Leftrightarrow p^3+q^3+r^3+6pqr \geq \sum_{\mathrm{sym}}p^2q \] which follows from Schur's Inequality ($p^3+q^3+r^3+3pqr\geq\sum_{\mathrm{sym}}p^2q$) and $pqr\geq0$. Equality holds at $p=q=0.5,q=0$ and permutations.

Prove that if $x,y,z>0$ and $x+y+z=1$ .then $$x^3+y^3+z^3+\frac{15}{4}xyz \geq \frac{1}{4},$$ $$x^2+y^2+z^2+\frac{9}{2}xyz \geq \frac{1}{2},$$ $$x^4+y^4+z^4+\frac{19}{8}xyz \geq \frac{1}{8}.$$