3. Determine all positive real numbers $a, b, c, d$ for which
1) $16abcd = \left(a^{2}+b^{2}+c^{2}+d^{2}\right)(a b c+b c d+c d a+d a b)$ and
2) $2 a b+2 c d+a c+a d+b c+b d=8$.
Solution. From condition 1) we have

$$
\begin{aligned}
1 & =\frac{16 a b c d}{\left(a^{2}+b^{2}+c^{2}+d^{2}\right)(a b c+b c d+c d a+d a b)}=\frac{1}{\frac{a^{2}+b^{2}+c^{2}+d^{2}}{4}} \cdot \frac{4}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}} \\
& \leq \frac{1}{\left(\frac{a+b+c+d}{4}\right)^{2}} \cdot \frac{a+b+c+d}{4}=\frac{4}{a+b+c+d}
\end{aligned}
$$

so $a+b+c+d \leq 4$. The equality holds if and only if $a=b=c=d=1$. From condition 2) we have

$$
\begin{aligned}
16 & =4 a b+4 c d+2 a c+2 a d+2 b c+2 b d \\
& =2 a b+2 c d+2 a b+2 c d+2 a c+2 a d+2 b c+2 b d \\
& \leq a^{2}+b^{2}+c^{2}+d^{2}+2 a b+2 c d+2 a c+2 a d+2 b c+2 b d \\
& =(a+b+c+d)^{2}
\end{aligned}
$$

so $4 \leq a+b+c+d$.

Therefore, $a+b+c+d=4$, from which it follows that $a=b=c=d=1$.