Solve for m

Given, $\frac{mx^2 + 3x + 4}{x^2 + 2x + 2} < 5$

$\implies mx^2+3x+4<5(x^2+2x+2)$

$\implies (m-5)x^2-7x-6<0$ , $\forall x \in R$

$\therefore$ Discriminant $< 0$

$\implies 49 + 24(m-5)<0$

$\implies m<\frac{71}{24}$

$\therefore m \leq 2$

$\begin{aligned} \frac {mx^2+3x+4}{x^2+2x+2} & < 5 \\ mx^2 + 3x + 4 & < 5x^2 + 10 x + 10 \\ (m-5)x^2 - 7x - 6 & < 0 \\ \implies 7^2 - 4(m-5)(-6) & < 0 \\ 24 m & < 71 \\ m & < \frac {71}{24} \approx 2.958333 \end{aligned}$

Therefore the maximum integer $m$ is $\boxed 2$ .