= Solution
Apply the volume <mean value property> to the two balls and use nonnegativity together with inclusion:
$$
u(w)=\frac1{|B(w,R)|}\int_{B(w,R)}u\ge\frac1{|B(w,R)|}\int_{B(z,r)}u=\boxed{\left(\frac rR\right)^3u(z).}
$$
For any $w,z\in B(x,r)$, the <triangle inequality> gives $B(z,r)\subset B(w,3r)\subset B(x,4r)\subset\Omega$. Taking $R=3r$ in the previous bound therefore gives $u(w)\ge u(z)/27$. Take the infimum over $w$ and supremum over $z$ to obtain
$$
\boxed{\inf_{B(x,r)}u\ge3^{-3}\sup_{B(x,r)}u.}
$$
No positivity stronger than nonnegativity is needed; if a value vanishes, this argument forces every value in the smaller ball to vanish too.
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