IMO Hk TST, 1997

The maximum possible value of $k$ is $4$.

Firstly, consider $a = 0$ and $b = c = \frac{1}{2}$. The inequality becomes
$$0 + \frac{2}{4+k} + \frac{2}{4+k} \ge \frac{1}{2}.$$
Hence, we need $k \le 4$.

It remains to prove
$$\frac{a}{1 + 9bc + 4(b - c)^2} + \frac{b}{1 + 9ca + 4(c - a)^2} + \frac{c}{1 + 9ab + 4(a - b)^2} \ge \frac{1}{2}$$
for all nonnegative $a, b, c$ with $a + b + c = 1$.

By the Cauchy-Schwarz inequality, we have
$$\left( \sum_{\text{cyc}} a(1 + 9bc + 4(b-c)^2) \right) \left( \sum_{\text{cyc}} \frac{a}{1 + 9bc + 4(b-c)^2} \right) \ge (a+b+c)^2 = 1.$$
So,
$$\sum_{\text{cyc}} \frac{a}{1 + 9bc + 4(b-c)^2} \ge \frac{1}{\sum_{\text{cyc}} a(1 + 9bc + 4(b-c)^2)}.$$
Therefore, it suffices to prove
$$\sum_{\text{cyc}} a(1 + 9bc + 4(b-c)^2) \le 2.$$
Expanding,
$$\sum_{\text{cyc}} a(1 + 9bc + 4(b-c)^2) = \sum_{\text{cyc}} a + 9\sum_{\text{cyc}} abc + 4\sum_{\text{cyc}} a(b-c)^2.$$
Since $a + b + c = 1$, $\sum_{\text{cyc}} a = 1$.

Now,
$$\sum_{\text{cyc}} a(b-c)^2 = \sum_{\text{cyc}} a(b^2 - 2bc + c^2) = \sum_{\text{cyc}} ab^2 + ac^2 - 2abc.$$
But $\sum_{\text{cyc}} ab^2 + ac^2 = \sum_{\text{sym}} a^2b$.

So,
$$\sum_{\text{cyc}} a(1 + 9bc + 4(b-c)^2) = 1 + 9abc + 4\sum_{\text{sym}} a^2b - 8abc.$$
Thus,
$$1 + 9abc + 4\sum_{\text{sym}} a^2b - 8abc = 1 + abc + 4\sum_{\text{sym}} a^2b.$$
Therefore, we need
$$1 + abc + 4\sum_{\text{sym}} a^2b \le 2.$$
Or,
$$abc + 4\sum_{\text{sym}} a^2b \le 1.$$
But $\sum_{\text{sym}} a^2b = a^2b + a^2c + b^2a + b^2c + c^2a + c^2b$.

Now, since $a + b + c = 1$, we can homogenize the expression:
$$(a + b + c)^3 \ge 3abc + 4(a^2b + a^2c + b^2a + b^2c + c^2a + c^2b).$$
Expanding $(a + b + c)^3$:
$$(a + b + c)^3 = a^3 + b^3 + c^3 + 3(a^2b + a^2c + b^2a + b^2c + c^2a + c^2b) + 6abc.$$
So,
$$a^3 + b^3 + c^3 + 3abc \ge a^2b + a^2c + b^2a + b^2c + c^2a + c^2b.$$
This is exactly Schur's inequality. So the inequality is proved.