Неравенство......................................

$\frac{ x^{2} }{3} \geq \frac{3x+3}{4}$

$\frac{ x^{2} }{3} - \frac{3x+3}{4} \geq 0$

$\frac{4 x^{2} -3(3x+3)}{12} \geq 0$

$\frac{4 x^{2} -9x-9}{12} \geq 0$ *12

${4 x^{2} -9x-9} \geq 0$

${4 x^{2} -9x-9} =0$

$D=(-9)^2-4*4*9=81+144=225$

$x_1= \frac{9+15}{8}=3$

$x_2= \frac{9-15}{8}=-0.75$

$4(x-3)(x+0.75) \geq 0$

Решаем методом интервалов:

-----------+---------[-0.75]------- - --------[3]---------+------------

Ответ: $(-$ ∞ $;-0.75]$ ∨ $[3;+$ ∞ $)$