(Iran TST 1996) Since the expressions are homogeneous, WLOG we may assume $ab + bc + ca = 1$. We need to minimize
$$f(a, b, c) = \frac{1}{(a+b)^2} + \frac{1}{(b+c)^2} + \frac{1}{(c+a)^2}.$$
WLOG assume $a \ge b \ge c$. Let $t$ be the positive root to $t^2 + 2ct = 1$, so that $(t)(t) + (t)(c) + (c)(t) = 1$, which means the triple $(t, t, c)$ also satisfies the constraint.
Claim. We have $f(a, b, c) \ge f(t, t, c)$.
Proof. Firstly, we have
$$f(t, t, c) = \frac{1}{4t^2} + \frac{2}{(t+c)^2} = \frac{1}{4t^2} + \frac{2}{t^2 + 2ct + c^2}.$$
For the latter term, observe that $t^2 + 2ct = 1 = ab + bc + ca$. Therefore, we obtain
$$\frac{2}{t^2 + 2ct + c^2} = \frac{2}{ab + bc + ca + c^2} = \frac{2}{(c+a)(c+b)}.$$
For the former term, we claim that $t^2 \ge ab$. Suppose on the contrary that $t^2 < ab$. Then we also have $a+b \ge 2\sqrt{ab} > 2t$. But this implies
$$ab + bc + ca = ab + c(a + b) > t^2 + 2ct,$$
which is a contradiction. Therefore, $t^2 \ge ab$, and hence
$$\frac{1}{4t^2} \le \frac{1}{4ab}.$$
Now, it suffices to show $f(a, b, c) \ge \frac{1}{4ab} + \frac{2}{(c+a)(c+b)}$. Indeed,
$$\begin{align*} f(a, b, c) &\ge \frac{1}{4ab} + \frac{2}{(c+a)(c+b)} \\ \Leftrightarrow \quad &\frac{1}{(b+c)^2} + \frac{1}{(c+a)^2} - \frac{2}{(c+a)(c+b)} \ge \frac{1}{4ab} - \frac{1}{(a+b)^2} \\ \Leftrightarrow \quad &\frac{((c+a) - (c+b))^2}{(c+a)^2(c+b)^2} \ge \frac{(a+b)^2 - 4ab}{4ab(a+b)^2} \\ \Leftrightarrow \quad &\frac{(a-b)^2}{(c+a)^2(c+b)^2} \ge \frac{(a-b)^2}{4ab(a+b)^2}. \end{align*}$$
This is true since
$$(c+a)^2(c+b)^2 \le (b+a)^2(b+b)^2 = 4b^2(a+b)^2 \le 4ab(a+b)^2$$
using the assumption $a \ge b \ge c$. This proves the claim. $\square$

By the claim, it remains to prove $f(t, t, c) \ge \frac{9}{4}$ for any positive real numbers $t$ and $c$ satisfying $t^2 + 2ct = 1$. Now, using $c = \frac{1-t^2}{2t}$,
$$\begin{align*} f(t, t, c) &\ge \frac{9}{4} \\ \Leftrightarrow \quad \frac{1}{4t^2} + \frac{2}{(t+c)^2} &\ge \frac{9}{4} \\ \Leftrightarrow \quad \frac{1}{4t^2} + \frac{8t^2}{(1+t^2)^2} &\ge \frac{9}{4} \\ \Leftrightarrow \quad \frac{-9t^6 + 15t^4 - 7t^2 + 1}{4t^2(1+t^2)^2} &\ge 0 \\ \Leftrightarrow \quad \frac{(1-t^2)(1-3t^2)^2}{4t^2(1+t^2)^2} &\ge 0. \end{align*}$$
As $1 = t^2 + 2ct > t^2$, this inequality is true. Equality holds when $t = \frac{1}{3}$, i.e. $a = b = c$. (Note that if we allow at most one $a$, $b$, $c$ to be $0$, equality also holds when $t = 1$, i.e. $a = b$ and $c = 0$ up to permutation.)