Philippine Mathematical Olympiad

Solution:

Since $(a-d)(b-c) \geq 0$ and $(a-b)(c-d) \geq 0$, then
$$ab + cd \geq ac + bd \geq ad + bc$$
From Equations (4) and (5), we have
$$\begin{gathered} (ab + cd) + (ac + bd) + (ad + bc) = \frac{1}{2}\left[(a+b+c+d)^{2} - \left(a^{2}+b^{2}+c^{2}+d^{2}\right)\right] = \\ \frac{1}{2}\left(13^{2} - 43\right) = 63 \end{gathered}$$
Thus, using Equation (6), we have $ab + cd \geq \frac{63}{3} = 21$. Since $c + d = 13 - (a + b)$, then
$$\begin{aligned} (a+b)^{2} + [13 - (a+b)]^{2} & = (a+b)^{2} + (c+d)^{2} \\ & = \left(a^{2} + b^{2} + c^{2} + d^{2}\right) + 2(ab + cd) \\ & \geq 43 + 2(21) = 85 \end{aligned}$$
Thus, we have
$$\begin{aligned} (a+b)^{2} + 169 - 26(a+b) + (a+b)^{2} & \geq 85 \\ (a+b)^{2} - 13(a+b) + 42 & \geq 0 \\ (a+b-6)(a+b-7) & \geq 0 \end{aligned}$$
which means either $a+b \leq 6$ or $a+b \geq 7$. However, since $2(a+b) \geq a+b+c+d = 13$, then $a+b \geq 6.5$. Thus, $a+b \geq 7$. From this, we have
$$\begin{aligned} (a+b)^{2} + (c-d)^{2} & \geq 7^{2} + 0^{2} \\ a^{2} + b^{2} + c^{2} + d^{2} + 2ab - 2cd & \geq 49 \\ 43 + 2(ab - cd) & \geq 49 \\ ab - cd & \geq 3 \\ ab & \geq 3 + cd \end{aligned}$$