The greatest value of $\frac{4}{4x^2+4x+9}$  is

$4x^{2}y+4xy+9y-4=0$

Where $y$  is real

$\therefore b^2-4ac\ge 0$

$y^2-9y^2+4y\ge 0$

$-8y^2+4y\ge 0$

$2y^2-y\le 0$

$\therefore y\left(2y-1\right)\le 0$

$y\in \left[0,\frac{1}{2}\right]$

$\therefore 0\le \frac{4}{4x^2+4x+9}\le \frac{1}{2}$