Suppose $a \geqslant b \geqslant c.$ $ \ $ There are two cases $\bf{1.}$ $ \ abc \leqslant 0 \Rightarrow a \geqslant b \geqslant 0 \geqslant c $ $$(1) \Leftrightarrow \sum \left(\frac{33a^2-a}{33a^2+1} + a \right) \geqslant 0 \Leftrightarrow \sum \frac{a^2 \left(a+1 \right)}{33a^2+1} \geqslant 0.$$Because $a,b \geqslant 0,$ $ \ $ we can apply the Cauchy - Schwarz inequality : $$\frac{a^2}{\frac{33a^2+1}{a+1}} + \frac{b^2}{\frac{33b^2+1}{b+1}} \geqslant \frac{\left(a+b \right)^2}{ \frac{33a^2+1}{a+1} + \frac{33b^2+1}{b+1}}=\frac{c^2}{\frac{33a^2+1}{a+1} + \frac{33b^2+1}{b+1}}.$$Therefore, it's enough to show $$\frac{1}{\frac{33a^2+1}{a+1} + \frac{33b^2+1}{b+1}} + \frac{c+1}{33c^2+1} \geqslant 0 \ \ \ \ (2)$$We prove $$\frac{33a^2+1}{a+1} + \frac{33b^2+1}{b+1} \leqslant \frac{33 \left(a+b \right)^2 + a+b+2}{a+b+1} \left(= \frac{33c^2-c+2}{1-c} \right).$$$$\Leftrightarrow \frac{34ab \left(a+b+2 \right)}{\left(a+1 \right) \left(b+1 \right) \left(a+b+1 \right)} \geqslant 0 \ \ (\text{OK}).$$$$\Rightarrow LHS_{(2)} \geqslant \frac{1-c}{33c^2-c+2} + \frac{c+1}{33c^2+1} =\frac{65c^2+3}{\left(33c^2-c+2 \right) \left(33c^2+1 \right)} > 0.$$$\bf{2.}$ $ \ abc > 0$ Denote by $\begin{cases} N = 33 a^2 - 33abc + 3a + 66bc - 2 \\ Q = 33b^2 - 33abc + 3b + 66ca - 2 \\ T = 33c^2 - 33abc + 3c + 66ab - 2 \end{cases}$ $ \Rightarrow N+Q+T = -99abc - 6 < 0 \ \left(\text{because} \ a+b+c=0 \right).$ So we can assume that $N < 0$ $$\Rightarrow f \left(a,b,c \right) \geqslant f \left(a, \frac{b+c}{2}, \frac{b+c}{2} \right) = \frac{99a^2 \left(33a^2+a+2 \right)}{\left(33a^2+1 \right) \left(33a^2+4 \right)} \geqslant 0,$$where $f \left(a,b,c \right) = \sum \frac{33a^2-a}{33a^2+1}.$

Similar to Balkan MO 2011/2.

@Doxuantrong Why we can know that \Rightarrow f \left(a,b,c \right) \geqslant f \left(a, \frac{b+c}{2}, \frac{b+c}{2} \right) ?