South-Afrika, 2011

Note that since $\tan 2\theta$ attains all real numbers in the interval $\left(-\frac{\pi}{4}, \frac{\pi}{4}\right)$, we may restrict our attention to this domain. Also,
$$f(-\tan 2\theta) = f(\tan 2(-\theta))$$
$$= \tan^4(-\theta) + \cot^4(-\theta) \\ = \tan^4 \theta + \cot^4 \theta \\ = f(\tan 2\theta),$$
so $f$ is an even function, and we may restrict $\theta$ to the interval $(0, \frac{\pi}{4})$.

Next, let $0 < x < y$, and let $x = \tan 2\alpha$, $y = \tan 2\beta$, where $0 < \alpha < \beta < \frac{\pi}{4}$. Then
$$\begin{aligned} f(x) - f(y) &= f(\tan 2\alpha) - f(\tan 2\beta) \\ &= \tan^4 \alpha + \cot^4 \alpha - \tan^4 \beta - \cot^4 \beta \\ &= (\tan^4 \alpha - \tan^4 \beta) - \left( \frac{1}{\tan^4 \beta} - \frac{1}{\tan^4 \alpha} \right) \\ &= (\tan^4 \alpha - \tan^4 \beta) \left( 1 - \frac{1}{\tan^4 \alpha \tan^4 \beta} \right). \end{aligned}$$
Now, $\tan \theta$ is increasing on the interval $(0, \frac{\pi}{4})$, and for $\theta \in (0, \frac{\pi}{4})$, $0 < \tan \theta < 1$, hence $\tan^4 \alpha - \tan^4 \beta < 0$ and $\tan^4 \alpha \tan^4 \beta < 1 \Rightarrow 1 - \frac{1}{\tan^4 \alpha \tan^4 \beta} < 0$, hence $f(x) - f(y) > 0$, showing that $f$ is decreasing.

Next, note that both $\cot \theta$ and $\tan \theta$ are concave up on the interval $(0, \frac{\pi}{4})$, so we have that $\frac{\tan 2\alpha + \tan 2\beta}{2} \ge \tan(\alpha + \beta)$, by Jensen's Inequality. Hence, setting $x = \tan 2\alpha$ and $y = \tan 2\beta$, and recalling that $f$ is decreasing, we have
$$\begin{aligned} f\left(\frac{x+y}{2}\right) &= f\left(\frac{\tan 2\alpha + \tan 2\beta}{2}\right) \\ &\le f\left(\tan 2\left(\frac{\alpha+\beta}{2}\right)\right) \\ &= \tan^4\left(\frac{\alpha+\beta}{2}\right) + \cot^4\left(\frac{\alpha+\beta}{2}\right) \\ &\le \frac{\tan^4\alpha + \tan^4\beta}{2} + \frac{\cot^4\alpha + \cot^4\beta}{2} \quad \text{(Jensen's Inequality)} \\ &= \frac{f(\tan 2\alpha) + f(\tan 2\beta)}{2} \\ &= \frac{f(x) + f(y)}{2}. \end{aligned}$$
This shows that $f$ is concave up.

Now let $x = \sin \alpha$ and $y = \cos \alpha$. Since $f$ is even, we may assume that both $x$ and $y$ are positive, and hence $\frac{x+y}{2} \le \sqrt{\frac{x^2+y^2}{2}}$. Using Jensen's Inequality and recalling that $f$ is decreasing, we have that
$$f(x) + f(y) \ge 2f\left(\frac{x+y}{2}\right) \ge 2f\left(\sqrt{\frac{x^2+y^2}{2}}\right) = 2f\left(\frac{1}{\sqrt{2}}\right).$$
Now, let $\tan 2\theta = \frac{1}{\sqrt{2}}$. Then $\tan \theta = \sqrt{3} - \sqrt{2}$ and $\cot \theta = \sqrt{3} + \sqrt{2}$, and so
$$\begin{aligned} f\left(\frac{1}{\sqrt{2}}\right) &= f(\tan 2\theta) \\ &= \tan^4 \theta + \cot^4 \theta \\ &= (\sqrt{3} - \sqrt{2})^4 + (\sqrt{3} + \sqrt{2})^4 \\ &= 2(\sqrt{3}^4 + 6(\sqrt{2}\sqrt{3})^2 + \sqrt{2}^4) \\ &= 98. \end{aligned}$$
Hence $f(\cos \theta) + f(\sin \theta) \ge 2(98) = 196$.