Let $a, b, c \geqslant 0, \sum_{\mathrm{cyc}} a=1$. Find the maximum value of $\sum_{\mathrm{cyc}} \sqrt{a^{2}+b c}$.

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When $b=c=\frac{1}{2}, a=0$; or $c=a=\frac{1}{2}, b=0$; or $a=b=\frac{1}{2}, c=0$,
$$\sum_{\mathrm{cyc}} \sqrt{a^{2}+b c}=\frac{3}{2} \sum_{\mathrm{cyc}} a$$

Now we prove that the maximum value of $\sum_{\mathrm{cyc}} \sqrt{a^{2}+b c}$ is $\frac{3}{2} \sum_{\mathrm{cyc}} a=\frac{3}{2}$.
Assume without loss of generality that $c \geqslant b \geqslant a \geqslant 0$, then
$$\sqrt{c^{2}+a b} \leqslant \sqrt{c^{2}+a c} \leqslant c+\frac{a}{2}$$

Additionally, we prove that
$$\sqrt{a^{2}+b c}+\sqrt{b^{2}+c a} \leqslant \frac{c}{2}+a+\frac{3 b}{2}$$

Because
$$(c-2 a-b)^{2}+8 a(b-a) \geqslant 0$$

So
$$\begin{array}{c}
c^{2}-(4 a+2 b) c-\left(4 a^{2}-b^{2}-12 a b\right) \geqslant 0 \\
2\left(a^{2}+b^{2}+b c+c a\right) \leqslant\left(\frac{c}{2}+a+\frac{3 b}{2}\right)^{2}
\end{array}$$

Also, because
$$\left(\sqrt{a^{2}+b c}+\sqrt{b^{2}+c a}\right)^{2} \leqslant 2\left(a^{2}+b^{2}+b c+c a\right)$$

So inequality (12) holds.
From inequalities (11) and (12), we know that $\sum_{\mathrm{cyc}} \sqrt{a^{2}+b c} \leqslant \frac{3}{2} \sum_{\mathrm{cyc}} a=\frac{3}{2}$.