Taiwan IMO Selection Camp

From $b+c > \sqrt{2}$ we obtain $b^2 + c^2 > 1$, so $a^2 = 3 - (b^2 + c^2) < 2$, that is $a < \sqrt{2} < b+c$.
Thus we have $b+c-a > 0$, and similarly $c+a-b > 0$ and $a+b-c > 0$.
We will use HÖLDER's inequality:
$$\frac{x_1^{p+1}}{y_1^p} + \frac{x_2^{p+1}}{y_2^p} + \cdots + \frac{x_n^{p+1}}{y_n^p} \ge \frac{(x_1 + x_2 + \cdots + x_n)^{p+1}}{(y_1 + y_2 + \cdots + y_n)^p}$$
which holds for all positive real numbers $p$, $x_1$, $x_2$, $\cdots$, $x_n$, $y_1$, $y_2$, $\cdots$, $y_n$. Applying HÖLDER's inequality with $p = 2$, $n = 3$ we obtain:
$$\begin{aligned} \sum \frac{a}{(b+c-a)^2} &= \sum \frac{(a^2)^3}{a^5 (b+c-a)^2} \\ &\ge \frac{(a^2 + b^2 + c^2)^3}{(\sum a^{5/2} (b+c-a))^2} \\ &= \frac{27}{(\sum a^{5/2} (b+c-a))^2}. \end{aligned}$$
To estimate the denominator on the right-hand side above, we observe a special case of SCHUR's inequality:
$$\sum a^{3/2} (a-b)(a-c) \ge 0$$
After rearranging, this gives:
$$\sum a^{5/2} (b+c-a) \le abc(\sqrt{a} + \sqrt{b} + \sqrt{c})$$
Then by the Cauchy-Schwarz inequality we know:
$$\left(\frac{\sqrt{a} + \sqrt{b} + \sqrt{c}}{3}\right)^4 \le \frac{a^2 + b^2 + c^2}{3} = 1$$
that is $\sqrt{a} + \sqrt{b} + \sqrt{c} \le 3$. Hence
$$\sum \frac{a}{(b+c-a)^2} \ge \frac{27}{(abc(\sqrt{a} + \sqrt{b} + \sqrt{c}))^2} \ge \frac{3}{(abc)^2}$$
as desired.