= Solution
Write $a=\min(C_1,C_2)>0$, $b=\max(C_1,C_2)$ and $c=C^0>0$. At crystal orientation $\theta$, the conductivity <tensor> is $Q_\theta\operatorname{diag}(a,b)Q_\theta^T$. Because the comparison <tensor> is scalar, its resolvent rotates in the same way:
$$
M_\theta=[I+(C_\theta-cI)/(2c)]^{-1}
=Q_\theta\operatorname{diag}\left(\frac{2c}{a+c},\frac{2c}{b+c}\right)Q_\theta^T.
$$
Uniform angular averaging uses $\langle\cos^2\theta\rangle=\langle\sin^2\theta\rangle=1/2$ and $\langle\sin\theta\cos\theta\rangle=0$. It therefore sends any such rotated diagonal <tensor> to half its trace times $I$. In particular,
$$
\langle M\rangle=c\left(\frac1{a+c}+\frac1{b+c}\right)I,
\qquad
\langle MC\rangle=c\left(\frac a{a+c}+\frac b{b+c}\right)I.
$$
Dividing these scalar coefficients in the stated <Hashin-Shtrikman conductivity variational principle> gives
$$
\boxed{C^{\mathrm{HS}}=H(c)I,\qquad H(c)=\frac{2ab+c(a+b)}{a+b+2c}.}
$$
This averaging calculation uses the uniform orientation model in the question; a nonuniform distribution need not give isotropic <effective conductivity>.
The lower <Hashin-Shtrikman conductivity bounds> use a comparison no larger than every local principal conductivity, so $0<c\le a$. The upper variational bound uses $c\ge b$. These opposite directions come from the sign of the contrast $C-cI$ in the polarization functional: positive contrast gives a maximizing lower principle, and negative contrast a minimizing upper principle. The endpoint values are understood as limits when a contrast eigenvalue vanishes. Since
$$
H'(c)=\frac{(a-b)^2}{(a+b+2c)^2}\ge0,
$$
the strongest lower and upper values in their admissible comparison ranges are, respectively,
$$
\boxed{C_l=H(a)=\frac{a(a+3b)}{3a+b},\qquad
C_u=H(b)=\frac{b(3a+b)}{a+3b}.}
$$
Thus $C_lI\preceq C^{\mathrm{eff}}\preceq C_uI$ in the statistically isotropic polycrystal setting of these <Hashin-Shtrikman conductivity bounds>.
The <self-consistent conductivity approximation> chooses the comparison conductivity equal to the predicted conductivity, $c=C_{\mathrm{SC}}=H(c)$. Multiplication by the positive denominator cancels the terms $c(a+b)$ and leaves $2c^2=2ab$. Only the positive root is an admissible conductivity:
$$
\boxed{C_{\mathrm{SC}}=\sqrt{ab},\qquad C^{\mathrm{SC}}=\sqrt{ab}\,I.}
$$
For a direct proof of its ordering relative to the variational bounds, set $t=\sqrt{b/a}\ge1$. Then
$$
\frac{C_{\mathrm{SC}}-C_l}{a}=\frac{(t-1)^3}{t^2+3},
\qquad
\frac{C_u-C_{\mathrm{SC}}}{a}=\frac{t(t-1)^3}{1+3t^2}.
$$
Both quantities are nonnegative, proving \b[$C_l\le C_{\mathrm{SC}}\le C_u$], with equality throughout when $a=b$. The <self-consistent conductivity approximation> is a closure calculation; no unrequested claim of exactness for every spatial arrangement follows just from this fixed-point computation.
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