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

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

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

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

Summing this inequality with three similar ones, we get

$$
\begin{aligned}
& \sqrt[4]{a^{2}+3 a b}+\sqrt[4]{b^{2}+3 b c}+\sqrt[4]{c^{2}+3 c d}+\sqrt[4]{d^{2}+3 d a} \leqslant \\
& \leqslant \frac{1}{2 \sqrt{2}}\left(\frac{5 a+3 b+8}{4}+\frac{5 b+3 c+8}{4}+\frac{5 c+3 d+8}{4}+\frac{5 d+3 a+8}{4}\right)= \\
& =\frac{1}{2 \sqrt{2}} \cdot \frac{8(a+b+c+d)+32}{4} \leqslant \frac{16}{2 \sqrt{2}}=4 \sqrt{2}
\end{aligned}
$$

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

Second solution. By the Cauchy-Bunyakovsky inequality for the sets of numbers $\sqrt[4]{a}, \sqrt[4]{b}$, $\sqrt[4]{c}, \sqrt[4]{d}$ and $\sqrt[4]{a+3 b}, \sqrt[4]{b+3 c}, \sqrt[4]{c+3 d}, \sqrt[4]{d+3 a}$, we have

$$
\begin{aligned}
& (\sqrt[4]{a(a+3 b)}+\sqrt[4]{b(b+3 c)}+\sqrt[4]{c(c+3 d)}+\sqrt[4]{d(d+3 a)})^{2} \leqslant \\
& \leqslant(\sqrt{a}+\sqrt{b}+\sqrt{c}+\sqrt{d})(\sqrt{a+3 b}+\sqrt{b+3 c}+\sqrt{c+3 d}+\sqrt{d+3 a})
\end{aligned}
$$

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 $\sqrt{a+3 b}, \sqrt{b+3 c}, \sqrt{c+3 d}, \sqrt{d+3 a}$ and $1,1,1,1$, we have

$$
\begin{aligned}
& (\sqrt{a+3 b}+\sqrt{b+3 c}+\sqrt{c+3 d}+\sqrt{d+3 a})^{2} \leqslant \\
& \quad \leqslant((a+3 b)+(b+3 c)+(c+3 d)+(d+3 a))(1+1+1+1)=16(a+b+c+a) \leqslant 64
\end{aligned}
$$

so $\sqrt{a+3 b}+\sqrt{b+3 c}+\sqrt{c+3 d}+\sqrt{d+3 a} \leqslant 8$. Therefore,

$$
\sqrt[4]{a^{2}+3 a b}+\sqrt[4]{b^{2}+3 b c}+\sqrt[4]{c^{2}+3 c d}+\sqrt[4]{d^{2}+3 d a} \leqslant \sqrt{4 \cdot 8}=4 \sqrt{2}
$$

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 $\sqrt[4]{a+3 b}, \sqrt[4]{b+3 c}, \sqrt[4]{c+3 d}, \sqrt[4]{d+3 a}$, we have

$$
\begin{aligned}
& (\sqrt[4]{a(a+3 b)}+\sqrt[4]{b(b+3 c)}+\sqrt[4]{c(c+3 d)}+\sqrt[4]{d(d+3 a)})^{2} \leqslant \\
& \leqslant(\sqrt{a}+\sqrt{b}+\sqrt{c}+\sqrt{d})(\sqrt{a+3 b}+\sqrt{b+3 c}+\sqrt{c+3 d}+\sqrt{d+3 a})
\end{aligned}
$$

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

$$
2 \sqrt{x+2 y}=\sqrt{(x+2 y) \cdot 4} \leqslant \frac{1}{2}(x+3 y+4)
$$

so

$$
\begin{aligned}
& \sqrt{a+3 b}+\sqrt{b+3 c}+\sqrt{c+3 d}+\sqrt{d+3 a} \leqslant \\
& \quad \leqslant \frac{a+3 b+4}{4}+\frac{b+3 c+4}{4}+\frac{c+3 d+4}{4}+\frac{d+3 a+4}{4}=\frac{4(a+b+c+d)+16}{4}=8
\end{aligned}
$$

Therefore,

$$
\sqrt[4]{a^{2}+3 a b}+\sqrt[4]{b^{2}+3 b c}+\sqrt[4]{c^{2}+3 c d}+\sqrt[4]{d^{2}+3 d a} \leqslant \sqrt{4 \cdot 8}=4 \sqrt{2}
$$

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