So which one is it?

This is a standard Cauchy Schwarz problem.

First, we will show that $(\sum \sqrt{ ab} )^2 \leq 3 \sum a^2$ . This is true because we have

$\begin{aligned} \frac{1}{2} (a^2 + a^2 + b^2 + c ^2) & \geq 2 a \sqrt{bc} \\ \frac{1}{2} (a^2 + b^2 + b^2 + c ^2) & \geq 2 b \sqrt{ca} \\ \frac{1}{2} (a^2 + b^2 + c^2 + c ^2) & \geq 2 c \sqrt{ab} \\ a^2 + b^2 + c^2 & \geq ab + bc + ca\\ \hline \\ 3(a^2 + b^2 + c^2 ) & \geq ( \sqrt{ab} + \sqrt{bc} + \sqrt{ca} ) ^ 2 \\ \end{aligned}$

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By Cauchy Schwarz, we have $\left ( \sum \frac{ ab}{ a^2 + 3b^2 } \right ) \left( \sum a^2 + 3b^2 \right ) \leq \left( \sum \sqrt{ab} \right) ^ 2$ , or that

$\sum \frac{ ab}{ a^2 + 3b^2 } \leq \frac{ (\sum \sqrt{ ab} ) ^ 2 } { 4 \sum a^2 } \leq \frac{3}{4}$

Thus, $4M + 4 = 7$ .