3. The sum of positive numbers $a, b, c$ and $d$ does not exceed 4. Find the maximum value of the expression

$$
\sqrt[4]{a^{2}(a+b)}+\sqrt[4]{b^{2}(b+c)}+\sqrt[4]{c^{2}(c+d)}+\sqrt[4]{d^{2}(d+a)}
$$
Answer: $4 \sqrt[4]{2}$

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

$$
\sqrt[4]{a^{2}(a+b)}=\frac{\sqrt[4]{2 a \cdot 2 a \cdot(a+b) \cdot 2}}{\sqrt[4]{8}} \leqslant \frac{1}{\sqrt[4]{8}} \cdot \frac{2 a+2 a+(a+b)+2}{4}=\frac{5 a+b+2}{4 \sqrt[4]{8}}
$$

Summing this inequality with three similar ones, we get

$$
\begin{aligned}
& \sqrt[4]{a^{2}(a+b)}+\sqrt[4]{b^{2}(b+c)}+\sqrt[4]{c^{2}(c+d)}+\sqrt[4]{d^{2}(d+a)} \leqslant \\
& \quad \leqslant \frac{1}{\sqrt[4]{8}}\left(\frac{5 a+b+2}{4}+\frac{5 b+c+2}{4}+\frac{5 c+d+2}{4}+\frac{5 d+a+2}{4}\right)= \\
& \quad=\frac{1}{\sqrt[4]{8}} \cdot \frac{6(a+b+c+d)+8}{4} \leqslant \frac{8}{\sqrt[4]{8}}=4 \sqrt[4]{2}
\end{aligned}
$$

Equality is achieved when $a=b=c=d=1$.

Second solution. By the Cauchy-Bunyakovsky inequality for the sets of numbers $\sqrt{a}, \sqrt{b}$, $\sqrt{c}, \sqrt{d}$ and $\sqrt[4]{a+b}, \sqrt[4]{b+c}, \sqrt[4]{c+d}, \sqrt[4]{d+a}$, we have

$$
\begin{aligned}
& \left(\sqrt[4]{a^{2}(a+b)}+\sqrt[4]{b^{2}(b+c)}+\sqrt[4]{c^{2}(c+d)}+\sqrt[4]{d^{2}(d+a)}\right)^{2} \leqslant \\
& \leqslant(a+b+c+d)(\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+d}+\sqrt{d+a}) \leqslant 4(\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+d}+\sqrt{d+a})
\end{aligned}
$$

And by the Cauchy-Bunyakovsky inequality for the sets $\sqrt{a+b}, \sqrt{b+c}, \sqrt{c+d}, \sqrt{d+a}$ and 1,1, 1,1, we have

$$
\begin{aligned}
& (\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+d}+\sqrt{d+a})^{2} \leqslant \\
& \quad \leqslant((a+b)+(b+c)+(c+d)+(d+a))(1+1+1+1)=8(a+b+c+d) \leqslant 32
\end{aligned}
$$

Therefore, $\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+d}+\sqrt{d+a} \leqslant 4 \sqrt{2}$. Thus,

$$
\sqrt[4]{a^{2}(a+b)}+\sqrt[4]{b^{2}(b+c)}+\sqrt[4]{c^{2}(c+d)}+\sqrt[4]{d^{2}(d+a)} \leqslant \sqrt{4 \cdot 4 \sqrt{2}}=4 \sqrt[4]{2}
$$

Equality is achieved when $a=b=c=d=1$.

Third solution. By the Cauchy-Bunyakovsky inequality for the sets of numbers $\sqrt{a}, \sqrt{b}$, $\sqrt{c}, \sqrt{d}$ and $\sqrt[4]{a+b}, \sqrt[4]{b+c}, \sqrt[4]{c+d}, \sqrt[4]{d+a}$, we have

$$
\begin{aligned}
& \left(\sqrt[4]{a^{2}(a+b)}+\sqrt[4]{b^{2}(b+c)}+\sqrt[4]{c^{2}(c+d)}+\sqrt[4]{d^{2}(d+a)}\right)^{2} \leqslant \\
& \leqslant(a+b+c+d)(\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+d}+\sqrt{d+a}) \leqslant 4(\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+d}+\sqrt{d+a})
\end{aligned}
$$

By the inequality of means for two numbers $\sqrt{2} \sqrt{x+y}=\sqrt{2(x+y)} \leqslant \frac{1}{2}(x+y+2)$, therefore

$$
\begin{aligned}
\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+d}+\sqrt{d+a} \leqslant \frac{a+b+2}{2 \sqrt{2}}+\frac{b+c+2}{2 \sqrt{2}} & +\frac{c+d+2}{2 \sqrt{2}}+\frac{d+a+2}{2 \sqrt{2}}= \\
= & \frac{2(a+b+c+d)+8}{2 \sqrt{2}} \leqslant 4 \sqrt{2}
\end{aligned}
$$

Therefore,

$$
\sqrt[4]{a^{2}(a+b)}+\sqrt[4]{b^{2}(b+c)}+\sqrt[4]{c^{2}(c+d)}+\sqrt[4]{d^{2}(d+a)} \leqslant \sqrt{4 \cdot 4 \sqrt{2}}=4 \sqrt[4]{2}
$$

Equality is achieved when $a=b=c=d=1$.