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

$$
\sqrt[4]{2 a^{2}+a^{2} b}+\sqrt[4]{2 b^{2}+b^{2} c}+\sqrt[4]{2 c^{2}+c^{2} d}+\sqrt[4]{2 d^{2}+d^{2} a}
$$
Answer: $4 \sqrt[4]{3}$

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

$$
\sqrt[4]{2 a^{2}+a^{2} b}=\frac{\sqrt[4]{3 a \cdot 3 a \cdot(2+b) \cdot 3}}{\sqrt[4]{27}} \leqslant \frac{1}{\sqrt[4]{27}} \cdot \frac{3 a+3 a+(2+b)+3}{4}=\frac{6 a+b+5}{4 \sqrt[4]{27}}
$$

Summing this inequality with three similar ones, we get

$$
\begin{aligned}
& \sqrt[4]{2 a^{2}+a^{2} b}+\sqrt[4]{2 b^{2}+b^{2} c}+\sqrt[4]{2 c^{2}+c^{2} d}+\sqrt[4]{2 d^{2}+d^{2} a} \leqslant \\
& \quad \leqslant \frac{1}{\sqrt[4]{27}}\left(\frac{6 a+b+5}{4}+\frac{6 b+c+5}{4}+\frac{6 c+d+5}{4}+\frac{6 d+a+5}{4}\right)= \\
& \quad=\frac{1}{\sqrt[4]{27}} \cdot \frac{7(a+b+c+d)+20}{4} \leqslant \frac{12}{\sqrt[4]{27}}=4 \sqrt[4]{3}
\end{aligned}
$$

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 $\sqrt[4]{2+b}, \sqrt[4]{2+c}, \sqrt[4]{2+d}, \sqrt[4]{2+a}$, we have

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

And by the Cauchy-Bunyakovsky inequality for the sets $\sqrt{2+a}, \sqrt{2+b}, \sqrt{2+c}, \sqrt{2+d}$ and 1,1,1,1, we have

$$
\begin{aligned}
& (\sqrt{2+a}+\sqrt{2+b}+\sqrt{2+c}+\sqrt{2+d})^{2} \leqslant \\
& \quad \leqslant((2+a)+(2+b)+(2+c)+(2+d))(1+1+1+1)=4(a+b+c+d+8) \leqslant 48
\end{aligned}
$$

Therefore, $\sqrt{2+a}+\sqrt{2+b}+\sqrt{2+c}+\sqrt{2+d} \leqslant 4 \sqrt{3}$. Thus,

$$
\sqrt[4]{2 a^{2}+a^{2} b}+\sqrt[4]{2 b^{2}+b^{2} c}+\sqrt[4]{2 c^{2}+c^{2} d}+\sqrt[4]{2 d^{2}+d^{2} a} \leqslant \sqrt{4 \cdot 4 \sqrt{3}}=4 \sqrt[4]{3}
$$

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 $\sqrt[4]{2+b}, \sqrt[4]{2+c}, \sqrt[4]{2+d}, \sqrt[4]{2+a}$, we have

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

By the inequality of means for two numbers $\sqrt{3} \sqrt{2+x}=\sqrt{3(2+x)} \leqslant \frac{1}{2}(2+x+3)=\frac{1}{2}(x+5)$, therefore

$$
\begin{aligned}
& \sqrt{2+a}+\sqrt{2+b}+\sqrt{2+c}+\sqrt{2+d} \leqslant \frac{a+5}{2 \sqrt{3}}+\frac{b+5}{2 \sqrt{3}}+\frac{c+5}{2 \sqrt{3}}+\frac{d+5}{2 \sqrt{3}}= \\
& =\frac{(a+b+c+d)+20}{2 \sqrt{3}} \leqslant 4 \sqrt{3}
\end{aligned}
$$

Thus,

$$
\sqrt[4]{2 a^{2}+a^{2} b}+\sqrt[4]{2 b^{2}+b^{2} c}+\sqrt[4]{2 c^{2}+c^{2} d}+\sqrt[4]{2 d^{2}+d^{2} a} \leqslant \sqrt{4 \cdot 4 \sqrt{3}}=4 \sqrt[4]{3}
$$

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