= Solution
For an orthogonal change of Cartesian coordinates $x'_i=Q_{ij}x_j$, an ordinary <rank>-$n$ <tensor> transforms by
$$
\boxed{T'_{i_1\cdots i_n}=Q_{i_1j_1}\cdots Q_{i_nj_n}T_{j_1\cdots j_n}.}
$$
An <isotropic tensor> is invariant under such rotations. Orthogonality gives $Q_{ip}Q_{jq}\delta_{pq}=\delta_{ij}$, so transforming any of the three products of two <Kronecker deltas> leaves it unchanged. Their linear combination with arbitrary scalar <coefficients> is therefore isotropic.
For completeness, these three products exhaust the <rank>-four isotropic <tensors> in three dimensions. Half-turns about coordinate axes force a nonzero component to contain each coordinate label an even number of times. Axis interchanges then leave just the values $X=c_{1111}$, $\alpha=c_{1122}$, $\beta=c_{1212}$ and $\gamma=c_{1221}$. Rotating the first two axes by $\pi/4$ gives $X=\tfrac12X+\tfrac12(\alpha+\beta+\gamma)$, so $X=\alpha+\beta+\gamma$. These component values are exactly those of the three-delta-product expression. Thus isotropy of the elastic medium gives this form for its elasticity <tensor>.
Contracting with the symmetric strain gives
$$
\sigma_{ij}=\alpha e_{kk}\delta_{ij}+\beta e_{ij}+\gamma e_{ji}
=\lambda e_{kk}\delta_{ij}+2\mu e_{ij},\qquad
\boxed{\lambda=\alpha,\quad2\mu=\beta+\gamma.}
$$
These are the <Lamé parameters>. Taking the <trace> determines the scalar part of the strain:
$$
\boxed{p=\frac13e_{kk},\qquad d_{ij}=e_{ij}-p\delta_{ij},\qquad d_{ii}=0.}
$$
Since $\sum_{ij}e_{ij}^2=3p^2+\sum_{ij}d_{ij}^2$, the stored <strain energy density> is
$$
E=\frac\lambda2(e_{kk})^2+\mu e_{ij}e_{ij}
=\frac32(3\lambda+2\mu)p^2+\mu\sum_{ij}d_{ij}^2.
$$
A nonzero traceless test strain proves necessity of $\mu\ge0$, and a pure scalar strain proves necessity of $3\lambda+2\mu\ge0$. Conversely these two conditions make both terms nonnegative for every strain. Thus <nonnegative isotropic elastic strain energy> is equivalent to
$$
\boxed{\mu\ge0,\qquad\lambda\ge-\frac23\mu.}
$$
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