3. The sum of positive numbers $a, b, c$ and $d$ is not less than 8. Find the minimum value of the expression

$\frac{a^{4}}{(a+b)(a+c)(a+d)}+\frac{b^{4}}{(b+c)(b+d)(b+a)}+\frac{c^{4}}{(c+d)(c+a)(c+b)}+\frac{d^{4}}{(d+a)(d+b)(d+c)}$.
Answer: 1

First solution. By the inequality of means for four numbers, we have

$$
\begin{aligned}
\frac{a^{4}}{(a+b)(a+c)(a+d)}+\frac{a+b}{16}+\frac{a+c}{16} & +\frac{a+d}{16} \geqslant \\
& \geqslant 4 \sqrt[4]{\frac{a^{4}}{(a+b)(a+c)(a+d)} \cdot \frac{a+b}{16} \cdot \frac{a+c}{16} \cdot \frac{a+d}{16}}=\frac{a}{2}
\end{aligned}
$$

Therefore,

$$
\frac{a^{4}}{(a+b)(a+c)(a+d)} \geqslant \frac{a}{2}-\left(\frac{a+b}{16}+\frac{a+c}{16}+\frac{a+d}{16}\right)=\frac{5 a}{16}-\frac{b+c+d}{16}
$$

Summing this inequality with three similar ones, we get

$$
\begin{aligned}
& \frac{a^{4}}{(a+b)(a+c)(a+d)}+\frac{b^{4}}{(b+c)(b+d)(b+a)}+\frac{c^{4}}{(c+d)(c+a)(c+b)}+\frac{d^{4}}{(d+a)(d+b)(d+c)} \geqslant \\
& \geqslant\left(\frac{5 a}{16}-\frac{b+c+d}{16}\right)+\left(\frac{5 b}{16}-\frac{c+d+a}{16}\right)+\left(\frac{5 c}{16}-\frac{d+a+b}{16}\right)+\left(\frac{5 d}{16}-\frac{a+b+c}{16}\right)= \\
& =\frac{2(a+b+c+d)}{16} \geqslant 1 .
\end{aligned}
$$

If $a=b=c=d=2$, then the sum of the fractions in the problem condition is 1, so the minimum value of the expression is 1.

Second solution. For brevity, let

$$
K=\frac{a^{4}}{(a+b)(a+c)(a+d)}+\frac{b^{4}}{(b+c)(b+d)(b+a)}+\frac{c^{4}}{(c+d)(c+a)(c+b)}+\frac{d^{4}}{(d+a)(d+b)(d+c)}
$$

By the Cauchy-Bunyakovsky inequality for the sets of numbers

$$
\frac{a^{4}}{(a+b)(a+c)(a+d)}, \frac{b^{4}}{(b+c)(b+d)(b+a)}, \frac{c^{4}}{(c+d)(c+a)(c+b)}, \frac{d^{4}}{(d+a)(d+b)(d+c)} \text { and }
$$

we have

$$
\begin{aligned}
& (2(a+b+c+d)) K=((a+d)+(b+a)+(c+b)+(d+c)) K \geqslant \\
& \quad \geqslant\left(\frac{a^{2}}{\sqrt{(a+b)(a+c)}}+\frac{b^{2}}{\sqrt{(b+c)(b+d)}}+\frac{c^{2}}{\sqrt{(c+d)(c+a)}}+\frac{d^{2}}{\sqrt{(d+a)(d+b)}}\right)^{2}
\end{aligned}
$$

Let the last expression in parentheses be denoted by $L$ and estimate it by the Cauchy-Bunyakovsky inequality for the sets

$$
\begin{aligned}
& \frac{a^{2}}{\sqrt{(a+b)(a+c)}}, \quad \frac{b^{2}}{\sqrt{(b+c)(b+d)}}, \quad \frac{c^{2}}{\sqrt{(c+d)(c+a)}}, \quad \frac{d^{2}}{\sqrt{(d+a)(d+b)}} \text { and } \\
& \sqrt{(a+b)(a+c)}, \quad \sqrt{(b+c)(b+d)}, \quad \sqrt{(c+d)(c+a)}, \quad \sqrt{(d+a)(d+b)} .
\end{aligned}
$$

Then $L M \geqslant(a+b+c+d)^{2}$, where

$$
M=\sqrt{(a+b)(a+c)}+\sqrt{(b+c)(b+d)}+\sqrt{(c+d)(c+a)}+\sqrt{(d+a)(d+b)}
$$

Thus,

$$
K \geqslant \frac{L^{2}}{2(a+b+c+d)} \geqslant \frac{(a+b+c+d)^{4}}{2(a+b+c+d) M^{2}}=\frac{(a+b+c+d)^{3}}{2 M^{2}}
$$

Finally, by the inequality of means for two numbers $\sqrt{(x+y)(x+z)} \leqslant \frac{(x+y)+(x+z)}{2}$, therefore

$$
M \leqslant \frac{2 a+b+c}{2}+\frac{2 b+c+d}{2}+\frac{2 c+d+a}{2}+\frac{2 d+a+b}{2}=2(a+b+c+d)
$$

Thus,

$$
K \geqslant \frac{1}{2} \cdot \frac{(a+b+c+d)^{3}}{(2(a+b+c+d))^{2}}=\frac{a+b+c+d}{8} \geqslant 1
$$

If $a=b=c=d=2$, then $K=1$, so the minimum value of $K$ is 1.

Third solution. For brevity, let

$$
K=\frac{a^{4}}{(a+b)(a+c)(a+d)}+\frac{b^{4}}{(b+c)(b+d)(b+a)}+\frac{c^{4}}{(c+d)(c+a)(c+b)}+\frac{d^{4}}{(d+a)(d+b)(d+c)}
$$

By the Cauchy-Bunyakovsky inequality for the sets of numbers

$$
\begin{aligned}
& \frac{a^{4}}{(a+b)(a+c)(a+d)}, \frac{b^{4}}{(b+c)(b+d)(b+a)}, \frac{c^{4}}{(c+d)(c+a)(c+b)}, \frac{d^{4}}{(d+a)(d+b)(d+c)} \text { and } \\
& (a+c)(a+d), \quad(b+d)(b+a), \quad(c+a)(c+b), \quad(d+b)(d+c)
\end{aligned}
$$

we have

$$
\begin{aligned}
(a+b+c+d)^{2} K & =((a+c)(a+d)+(b+d)(b+a)+(c+a)(c+b)+(d+b)(d+c)) K \geqslant \\
& \geqslant\left(\frac{a^{2}}{\sqrt{a+b}}+\frac{b^{2}}{\sqrt{b+c}}+\frac{c^{2}}{\sqrt{c+d}}+\frac{d^{2}}{\sqrt{d+a}}\right)^{2}
\end{aligned}
$$

Let the last expression in parentheses be denoted by $L$ and estimate it by the Cauchy-Bunyakovsky inequality for the sets

$$
\frac{a^{2}}{\sqrt{a+b}}, \quad \frac{b^{2}}{\sqrt{b+c}}, \quad \frac{c^{2}}{\sqrt{c+d}}, \quad \frac{d^{2}}{\sqrt{d+a}} \text { and } \sqrt{a+b}, \sqrt{b+c}, \sqrt{c+d}, \sqrt{d+a}
$$

Then $L M \geqslant(a+b+c+d)^{2}$, where $M=\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+d}+\sqrt{d+a}$. Thus,

$$
K \geqslant \frac{L^{2}}{(a+b+c+d)^{2}} \geqslant \frac{1}{(a+b+c+d)^{2}} \cdot \frac{(a+b+c+d)^{4}}{M^{2}}=\left(\frac{a+b+c+d}{M}\right)^{2}
$$

By the inequality of means for two numbers $2 \sqrt{x+y} \leqslant \frac{x+y+4}{2}$, therefore

$$
M \leqslant \frac{a+b+4}{4}+\frac{b+c+4}{4}+\frac{c+d+4}{4}+\frac{d+a+4}{4}=\frac{2 s+16}{4}
$$

where $s=a+b+c+d \geqslant 8$. Thus, we have proved that

$$
K \geqslant\left(\frac{s}{M}\right)^{2} \geqslant\left(\frac{4 s}{2 s+16}\right)^{2}
$$

It remains to note that $4 s=2 s+2 s \geqslant 2 s+16$, so the last expression is not less than 1. If $a=b=c=d=2$, then $K=1$, so the minimum value of $K$ is 1.