author observed in [@Soi:CAB15], for an edge-regular graph with parameters $(v,k,\lambda)$ we have $C_\Gamma(0, y) = -y(y - 1)(y-(\lambda + 2))$, so for all $y > \lambda+2$, we have $C_\Gamma(0,y)<0$. Hence the clique adjacency bound is always at most the trivial bound of $\lambda+2$. Therefore, to prove Theorem \[thm:limitation\], it suffices to show that, for strongly regular graph parameters satisfying $\lambda + 1 \leqslant -k/s$, the clique adjacency bound is at least $\lambda+2$.

\[lem:lam1le\] Let $\Gamma$ be a connected type-II strongly regular graph with parameters $(v,k,\lambda,\mu)$ and eigenvalues $k > r > s$. Suppose that $\lambda + 1 \leqslant -k/s$. Then $C_\Gamma(1, \lambda+2) \geqslant 0$ with equality if and only if $\lambda = -k/s - 1$.

First suppose $\lambda + 1 = -k/s$. Equivalently, since $\lambda + 1 = k + (r+1)(s+1)$ and $\mu = k + rs$, we have $-\mu/s = 1$. In this case, $C_\Gamma(1, \lambda+2) = C_\Gamma(-\mu/s, 1-k/s)$, which is zero by Equation .

It remains to assume $\lambda+1 < -k/s$. Using Proposition \[pro:SRG\] (see Appendix \[sec:comp\]) we can write $$\frac{\mu}{2} C_\Gamma(1,\lambda+2) = k(k-(\mu+1)(\lambda+1))+\mu(\lambda+1)^2.$$ To show this quantity is nonnegative, it suffices to show that $k-(\mu+1)(\lambda+1)$ is nonnegative. Using the inequality $\lambda+1 < -k/s$, we have $k-(\mu+1)(\lambda+1) 