= Solution
At the origin, the <Poisson kernel> is one, so the <mean value property for harmonic functions> gives
$$
u(0)=\frac1{2\pi}\int_0^{2\pi}f(\phi')\,d\phi'.
$$
For $0<r<1$, the denominator in the <Poisson kernel> satisfies $1-2r\cos\vartheta+r^2\geq(1-r)^2$. Thus
$$
0<P(r,\vartheta)\leq\frac{1-r^2}{(1-r)^2}=\frac{1+r}{1-r}.
$$
The nonnegativity of $f$ allows this upper bound to be inserted under the <Poisson integral on the unit disk>:
$$
\boxed{u(r,\phi)\leq\frac{1+r}{1-r}\,u(0)}.
$$
This is the upper bound in the <Harnack inequality on the unit disk>. The same argument with the minimum of the <Poisson kernel> gives the corresponding lower bound $(1-r)u(0)/(1+r)$.
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