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

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

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

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

Summing this inequality with three similar ones, we get

$$
\begin{aligned}
& \sqrt[4]{a(b+2 c)}+\sqrt[4]{b(c+2 d)}+\sqrt[4]{c(d+2 a)}+\sqrt[4]{d(a+2 b)} \leqslant \\
& \quad \leqslant \frac{1}{\sqrt[4]{3^{3}}}\left(\frac{3 a+b+2 c+6}{4}+\frac{3 b+c+2 d+6}{4}+\frac{3 c+d+2 a+6}{4}+\frac{3 d+a+2 b+6}{4}\right)= \\
& \quad=\frac{1}{\sqrt[4]{3^{3}}} \cdot \frac{6(a+b+c+d)+24}{4} \leqslant \frac{12}{\sqrt[4]{3^{3}}}=4 \sqrt[4]{3}
\end{aligned}
$$

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

Second solution. By the Cauchy-Bunyakovsky inequality for the sets of numbers $\sqrt[4]{a}, \sqrt[4]{b}$, $\sqrt[4]{c}, \sqrt[4]{d}$ and $\sqrt[4]{b+2 c}, \sqrt[4]{c+2 d}, \sqrt[4]{d+2 a}, \sqrt[4]{a+2 b}$, we have

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

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

$$
(\sqrt{a}+\sqrt{b}+\sqrt{c}+\sqrt{d})^{2} \leqslant(a+b+c+d)(1+1+1+1) \leqslant 4^{2}
$$

so $\sqrt{a}+\sqrt{b}+\sqrt{c}+\sqrt{d} \leqslant 4$. Similarly, by the Cauchy-Bunyakovsky inequality for the sets $\sqrt{b+2 c}, \sqrt{c+2 d}, \sqrt{d+2 a}, \sqrt{a+2 b}$ and $1,1,1,1$, we have

$$
\begin{aligned}
& (\sqrt{b+2 c}+\sqrt{c+2 d}+\sqrt{d+2 a}+\sqrt{a+2 b})^{2} \leqslant \\
& \quad \leqslant((b+2 c)+(c+2 d)+(d+2 a)+(a+2 b))(1+1+1+1)=12(a+b+c+a) \leqslant 48
\end{aligned}
$$

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

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

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

Third solution. By the Cauchy-Bunyakovsky inequality for the sets of numbers $\sqrt[4]{a}, \sqrt[4]{b}$, $\sqrt[4]{c}, \sqrt[4]{d}$ and $\sqrt[4]{b+2 c}, \sqrt[4]{c+2 d}, \sqrt[4]{d+2 a}, \sqrt[4]{a+2 b}$, we have

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

We estimate the factors on the right side separately. By the inequality of means for two numbers $\sqrt{x} \leqslant \frac{1}{2}(x+1)$, so

$$
\sqrt{a}+\sqrt{b}+\sqrt{c}+\sqrt{d} \leqslant \frac{a+1}{2}+\frac{b+1}{2}+\frac{c+1}{2}+\frac{d+1}{2}=\frac{a+b+c+d+4}{2} \leqslant 4
$$

Similarly, by the inequality of means for two numbers

$$
\sqrt{3} \sqrt{x+2 y}=\sqrt{(x+2 y) \cdot 3} \leqslant \frac{1}{2}(x+2 y+3)
$$

so

$$
\begin{aligned}
& \sqrt{b+2 c}+\sqrt{c+2 d}+\sqrt{d+2 a}+\sqrt{a+2 b} \leqslant \\
& \quad \leqslant \frac{b+2 c+3}{2 \sqrt{3}}+\frac{c+2 d+3}{2 \sqrt{3}}+\frac{d+2 a+3}{2 \sqrt{3}}+\frac{a+2 b+3}{2 \sqrt{3}}=\frac{3(a+b+c+d)+12}{2 \sqrt{3}}=4 \sqrt{3} .
\end{aligned}
$$

Therefore,

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

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