In Section 3 of https://arxiv.org/pdf/0708.3990.pdf Soundararajan shows that there exists infinitely many fundamental discriminants $8d$ for which $L(1/2, \chi_{8d})$ is very small. His argument runs as follows : Define, $$ M_1(R; X) := \sum_{X/16 \leq d \leq X/8} \mu(2d)^2 R(8d)^2 $$ and $$ M_2(R; X) := \sum_{X/16 \leq d \leq X/8} \mu(2d)^2 L(\tfrac 12, \chi_{8d}) R(8d)^2 $$ where $R$ are some real-valued weights. On the beginning of p. 11, Soundararajan shows that a suitable choice of $R$ leads to the bound, $$ \frac{M_2(R; X)}{M_1(R; X)} \ll \exp \Big ( - \Big ( \frac{1}{\sqrt{5}} + o(1) \Big ) \frac{\sqrt{\log X}}{\log\log X} \Big ) $$ and concludes that this implies $$ L(\tfrac 12, \chi_{8d}) \ll \exp \Big ( - \Big ( \frac{1}{\sqrt{5}} + o(1) \Big ) \frac{\sqrt{\log X}}{\log\log X} \Big ). $$ I do not understand this implication : If one assumes that $L(\tfrac 12, \chi_{8d}) \geq 0$ then it is certainly correct, but this is not known (although expected) to be true. In particular if $L(\tfrac 12, \chi_{8d})$ exhibits sign cancellations then I don't see how the previous displayed equation leads to the conclusion about small values of $L(1/2, \chi_{8d})$.

This can be corrected by working with $L(\tfrac 12, \chi_{8d})^2$ instead of $L(\tfrac 12, \chi_{8d})$ but leads to a worse numerical value than $\tfrac{1}{\sqrt{5}}$.

I would like to know if this is an oversight in Soundararajan's paper or my misunderstanding.