= Solution
For the lower variational inequality choose reference conductivity $a_1$ and a trial polarization $q$ that vanishes in phase 1. Let $v_q\in H_0^1(\Omega)$ solve
$$
a_1\Delta v_q+\nabla\cdot q=0.
$$
The <Hashin-Shtrikman conductivity variational principle> is
$$
\boxed{
J(u)\ge a_1|\Omega||\lambda|^2
+2\lambda\cdot\int_\Omega q\,dx
+\int_\Omega q\cdot\nabla v_q\,dx
-\int_{\mathrm{phase}\ 2}q\cdot(a-a_1I)^{-1}q\,dx .
}
$$
It holds for every admissible trial $q$; the exact principle takes the supremum of the right side. Writing the last integral only over phase 2 avoids applying an inverse to the zero contrast in the reference phase.
With $D=a_2-a_1>0$, the <isotropic> phase trial $q=tD\lambda\chi_2$ gives the lower expression
$$
a_1+Dp_2\left[2t-t^2\left(1+\frac{Dp_1}{3a_1}\right)\right].
$$
Here the quadratic is concave, so maximize it at $t=[1+Dp_1/(3a_1)]^{-1}$. The lower member of the <Hashin-Shtrikman conductivity bounds> is
$$
\boxed{
a^*\ge a_1+\frac{3a_1(a_2-a_1)p_2}{3a_1+p_1(a_2-a_1)}.
}
$$
Together with part (iii), this gives the required upper and lower interval. All denominators are positive for positive phase conductivities.
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