11. Given positive real numbers $a, b, c$ satisfying $a b+b c+c a=1$, find the minimum value of $\sqrt{a^{2}+a b+b^{2}}+\sqrt{b^{2}+b c+c^{2}}+\sqrt{c^{2}+c a+a^{2}}$.
Solution
Analysis 1: Since $a^{2}+a b+b^{2}=(a+b)^{2}-a b \geqslant(a+b)^{2}-\frac{1}{4}(a+b)^{2}=\frac{3}{4}(a+b)^{2}$,
then $\sqrt{a^{2}+a b+b^{2}}+\sqrt{b^{2}+b c+c^{2}}+\sqrt{c^{2}+c a+a^{2}} \geqslant \frac{\sqrt{3}}{2}(a+b+b+c+c+a)$ $=\sqrt{3}(a+b+c)$, and $(a+b+c)^{2} \geqslant 3(a b+b c+c a)=3 \Rightarrow a+b+c \geqslant \sqrt{3}$,
so the minimum value of $\sqrt{a^{2}+a b+b^{2}}+\sqrt{b^{2}+b c+c^{2}}+\sqrt{c^{2}+c a+a^{2}}$ is 3, with equality when $a=b=c=\frac{\sqrt{3}}{3}$.
Analysis 2: As shown in the figure, construct $P A=a, P B=b, P C=c$,
and the angles between $P A, P B, P C$ are $120^{\circ}$, according to the problem,
$S_{\triangle A B C}=\frac{1}{2}(a b+b c+c a) \sin 120^{\circ}=\frac{\sqrt{3}}{4}$ is a constant,
find the minimum value of $x+y+z$.
Let $x+y+z=2 p$, by Heron's formula we get
$$
\frac{\sqrt{3}}{4}=\sqrt{p(p-x)(p-y)(p-z)} \leqslant \sqrt{p \cdot\left(\frac{p-x+p-y+p-z}{3}\right)^{3}}=\sqrt{\frac{p^{4}}{27}}=\frac{\sqrt{3} p^{2}}{9} \Rightarrow p \geqslant \frac{3}{2} \text {, }
$$

equality holds when $x=y=z \Rightarrow a=b=c=\frac{\sqrt{3}}{3}$. Therefore, the minimum value of $x+y+z$ is 3.