Problem 1059

Notice the original inequality is equivalent to
$$\sum_{cyc} \left( \frac{33a^2 - a}{33a^2 + 1} + t \right) \ge 3t \quad (1)$$
We want to try to make every term on the left hand side non negative, so we need to choose a $t$ such that the numerator of $\left(\frac{33a^2-a}{33a^2+1}+t\right)$ is always positive.
Hence we want
$$\left( \frac{33a^2 - a}{33a^2 + 1} + t \right) = \frac{(1+t)33a^2 - a + t}{33a^2 + 1} = \frac{\left(\sqrt{33(1+t)}a - \sqrt{t}\right)^2}{33a^2 + 1} \quad (2)$$
to hold. Compare the coefficients and we need
$$1 = 2\sqrt{33t}\sqrt{1+t} \implies t = \frac{-1}{2} + \sqrt{\frac{17}{66}} > 0.$$
This number itself isn't really important, we just need to prove that it exists and $0 \le t \le 1$. For simplicity we will keep on using $t$ to denote $\frac{-1}{2} + \sqrt{\frac{17}{66}}$. Now we need to proof Eq. (1) holds, using the formula Eq. (2) we will get the original problem is equivalent to the following inequality
$$\sum_{cyc} \frac{(\sqrt{33(1+t)}a - \sqrt{t})^2}{33a^2 + 1} \ge 3t.$$
The denominator is annoying to deal with, so we want to try to make it simpler using our condition, that is
$$a^2 = (0-b-c)^2 = (b+c)^2 \le 2b^2+2c^2 \implies 3a^2 = a^2+2a^2 \le 2(a^2+b^2+c^2).$$
Similarly
$$3b^2 \le 2(a^2 + b^2 + c^2), \quad 3c^2 \le 2(a^2 + b^2 + c^2)$$
also holds. Therefore we have
$$\begin{align*} \sum_{cyc} \frac{(\sqrt{33(1+t)}a - \sqrt{t})^2}{33a^2 + 1} &\ge \sum_{cyc} \frac{(\sqrt{33(1+t)}a - \sqrt{t})^2}{22(a^2 + b^2 + c^2) + 1} \\ &= \frac{\sum_{cyc} (\sqrt{33(1+t)}a - \sqrt{t})^2}{22(a^2 + b^2 + c^2) + 1} \\ &= \frac{\sum_{cyc} (33(1+t)a^2) + 3t}{22(a^2 + b^2 + c^2) + 1} \end{align*}$$
Finally we need to prove it is greater than $3t$, which is just
$$\frac{\sum_{cyc} (33(1+t)a^2) + 3t}{22(a^2 + b^2 + c^2) + 1} \ge 3t \Leftrightarrow 33(1+t)(a^2 + b^2 + c^2) \ge 22(a^2 + b^2 + c^2)3t$$
Which holds if and only if $t \le 1$, therefore we proved the inequality.

Another Solution: If $a, b, c \ge -1$, then
$$\begin{aligned} \frac{33a^2 - a}{33a^2 + 1} &\ge -a \\ \Leftrightarrow 33a^2 - a &\ge -33a^3 - a \\ \Leftrightarrow 33a^3 + 33a^2 &\ge 0 \\ \Leftrightarrow 33a^2(a + 1) &\ge 0, \end{aligned}$$
which is true for $a, b, c$. Therefore
$$\sum_{cyc} \frac{33a^2 - a}{33a^2 + 1} \ge \sum_{cyc} (-a) = 0.$$
Now if one of $a, b, c$ is smaller than $(-1)$, suppose that $a < -1$. The inequality that we want to prove is equivalent to
$$\sum_{cyc} \frac{a+1}{33a^2+1} \le 3.$$
For any $x \in \mathbb{R}$, if
$$k = \frac{x+1}{33x^2+1},$$
then $33kx^2 - x + (k-1) = 0$. Therefore the determinant
$$\Delta = 1 - 132k(k-1) \ge 0.$$
This implies that
$$132k^2 - 132k - 1 \le 0,$$
and so
$$k \le \frac{1 + \sqrt{1 + \frac{1}{33}}}{2} < \frac{3}{2}.$$
Therefore,
$$\frac{a+1}{33a^2+1} + \frac{b+1}{33b^2+1} + \frac{c+1}{33c^2+1} < 0 + \frac{3}{2} + \frac{3}{2} = 3.$$
In conclusion, the inequality holds in both case, and so we are done.