2. Find the minimum value of the expression for $a, b>0$

$$
\frac{|6 a-4 b|+|3(a+b \sqrt{3})+2(a \sqrt{3}-b)|}{\sqrt{a^{2}+b^{2}}}
$$
Answer: $\sqrt{39}$.

Solution 1. The expression to be minimized is $2\left(d_{1}+d_{2}\right)$, where $d_{1}, d_{2}$ are the distances from point $A(3,2)$ to the lines $\ell_{1}, \ell_{2}$, given by the equations

$$
\ell_{1}: a x-b y=0 \quad \text { and } \quad \ell_{2}:(a+b \sqrt{3}) x+(a \sqrt{3}-b) y=0 .
$$

These lines intersect at the origin $O$. Let's find the angle $\varphi$ between them:

$$
\cos \varphi=\frac{a(a+b \sqrt{3})-b(a \sqrt{3}-b)}{\sqrt{a^{2}+b^{2}} \cdot \sqrt{(a+b \sqrt{3})^{2}+(a \sqrt{3}-b)^{2}}}=\frac{a^{2}+b^{2}}{\sqrt{a^{2}+b^{2}} \cdot \sqrt{4 a^{2}+4 b^{2}}}=\frac{1}{2}
$$

that is, $\varphi=\frac{\pi}{3}$. Without loss of generality, we assume that point $A$ lies inside the acute angle formed by the lines $\ell_{1}$ and $\ell_{2}$ (otherwise, we can align the lines $\ell_{1}$ and $A B$, reducing $d_{1}$ and $d_{2}$). Let $\alpha$ be the angle between $\ell_{1}$ and $A B$. Then $\alpha \in\left[0, \frac{\pi}{3}\right]$ and

$$
2\left(d_{1}+d_{2}\right)=2 \cdot O A \cdot\left(\sin \alpha+\sin \left(\frac{\pi}{3}-\alpha\right)\right)=4 \sqrt{13} \sin \frac{\pi}{6} \cos \left(\frac{\pi}{6}-\alpha\right)=2 \sqrt{13} \cos \left(\frac{\pi}{6}-\alpha\right)
$$

The minimum of the right-hand side is achieved when $\alpha=0$, that is, in the case $A \in \ell_{1}$, and it is equal to $\sqrt{39}$.

Solution 2. Let the minimized expression be denoted by $A$. Suppose $a=r \cos t, b=r \sin t$, where $r>0$, $t \in\left(0, \frac{\pi}{2}\right)$. Then

$$
\begin{aligned}
A & =|6 \cos t-4 \sin t|+|3(\cos t+\sqrt{3} \sin t)+2(\sqrt{3} \cos t-\sin t)|= \\
& =|6 \cos t-4 \sin t|+\left|6 \cos \left(t-\frac{\pi}{3}\right)-4 \sin \left(t-\frac{\pi}{3}\right)\right|=2 \sqrt{13}\left(|\sin (t-\alpha)|+\left|\sin \left(t-\frac{\pi}{3}-\alpha\right)\right|\right)
\end{aligned}
$$

where $\alpha=\operatorname{arctg} \frac{3}{2}$. Note that $\alpha \in\left(\frac{\pi}{6}, \frac{\pi}{3}\right)$, since $\frac{3}{2}\alpha$. Then

$$
A=2 \sqrt{13}\left(\sin (t-\alpha)+\sin \left(\frac{\pi}{3}+\alpha-t\right)\right)=4 \sqrt{13} \sin \frac{\pi}{6} \cos \left(t-\alpha-\frac{\pi}{6}\right)>2 \sqrt{13} \cos \frac{\pi}{6}=\sqrt{39}
$$

since $t-\alpha-\frac{\pi}{6}>-\frac{\pi}{6}$ and $t-\alpha-\frac{\pi}{6}<\frac{\pi}{2}-\frac{\pi}{6}-\alpha<\frac{\pi}{3}-\frac{\pi}{6}=\frac{\pi}{6}$. Thus, the minimum of $A$ is realized in case 1$)$.