Determine all quadruples $(a, b, c, d)$ of positive real numbers that satisfy $a+b+c+d=1$ and

$$
\max \left(\frac{a^{2}}{b}, \frac{b^{2}}{a}\right) \cdot \max \left(\frac{c^{2}}{d}, \frac{d^{2}}{c}\right)=(\min (a+b, c+d))^{4}
$$
If $a \geq b$ then $\frac{a}{b} \geq \frac{b}{a}$ and if just $a \leq b$ then $\frac{a}{b} \leq \frac{b}{a}$, so the sequence $(a, b)$ and the sequence $\left(\frac{a}{b}, \frac{b}{a}\right)$ are equally ordered. The rearrangement inequality now says that

$$
a \cdot \frac{a}{b}+b \cdot \frac{b}{a} \geq a \cdot \frac{b}{a}+b \cdot \frac{a}{b}
$$

We can rewrite this as $\frac{a^{2}}{b}+\frac{b^{2}}{a} \geq a+b$. If we combine this with $\max \left(\frac{a^{2}}{b}, \frac{b^{2}}{a}\right) \geq$ $\frac{\frac{a^{2}}{b}+\frac{b^{2}}{a}}{2}$, then we see

$$
\max \left(\frac{a^{2}}{b}, \frac{b^{2}}{a}\right) \geq \frac{a+b}{2} .
$$

Similarly, we also have

$$
\max \left(\frac{c^{2}}{d}, \frac{d^{2}}{d}\right) \geq \frac{c+d}{2}
$$

so

$$
\max \left(\frac{a^{2}}{b}, \frac{b^{2}}{a}\right) \cdot \max \left(\frac{c^{2}}{d}, \frac{d^{2}}{c}\right) \geq \frac{(a+b)(c+d)}{4}
$$

Without loss of generality, we assume that $\min (a+b, c+d)=a+b$. Since $a+b+c+d=1$, we have $a+b \leq \frac{1}{2}$. Therefore,

$$
\frac{(a+b)(c+d)}{4} \geq \frac{1}{4} \cdot(a+b)^{2} \geq(a+b)^{2} \cdot(a+b)^{2}=(a+b)^{4}
$$

We conclude that

$$
\max \left(\frac{a^{2}}{b}, \frac{b^{2}}{a}\right) \cdot \max \left(\frac{c^{2}}{d}, \frac{d^{2}}{c}\right) \geq(a+b)^{4}
$$

Given that equality holds here, equality must hold everywhere in the above steps. In particular, $a+b=\frac{1}{2}$ and $\max \left(\frac{a^{2}}{b}, \frac{b^{2}}{a}\right)=\frac{\frac{a^{2}}{b}+\frac{b^{2}}{a}}{2}$, so $\frac{a^{2}}{b}=\frac{b^{2}}{a}$, which implies $a^{3}=b^{3}$, hence $a=b$. This means that $a=b=\frac{1}{4}$. Since $c+d=1-(a+b)=\frac{1}{2}$, we can similarly deduce that $c=d=\frac{1}{4}$.
The only quadruple that can satisfy this is thus $(a, b, c, d)=\left(\frac{1}{4}, \frac{1}{4}, \frac{1}{4}, \frac{1}{4}\right)$. Substituting this into the given equality shows that both sides are $\frac{1}{16}$, so this quadruple satisfies the equality.