 Solution

For the lower variational inequality choose reference conductivity $a_1$ and a trial polarization $q$ that vanishes in phase 1. Let $v_q\in H_0^1(\Omega )$ solveThe Hashin-Shtrikman conductivity variational principle isIt 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 expressionHere the quadratic is concave, so maximize it at $t=[1+D{p}_1/(3a_1){]}^{-1}$. The lower member of the Hashin-Shtrikman conductivity bounds isTogether with part (iii), this gives the required upper and lower interval. All denominators are positive for positive phase conductivities.