4. Let positive real numbers $a, b, c$ satisfy $\frac{2}{a}+\frac{1}{b}=\frac{\sqrt{3}}{c}$. Then the minimum value of $\frac{2 a^{2}+b^{2}}{c^{2}}$ is $\qquad$ .
4. 9 .

From the given, we have $\frac{2 c}{a}+\frac{c}{b}=\sqrt{3}$.
By the Cauchy-Schwarz inequality and the AM-GM inequality, we get
$$
\begin{array}{l}
\frac{2 a^{2}+b^{2}}{c^{2}} \\
=\frac{1}{3}(2+1)\left[2\left(\frac{a}{c}\right)^{2}+\left(\frac{b}{c}\right)^{2}\right] \\
\geqslant \frac{1}{3}\left(\frac{2 a}{c}+\frac{b}{c}\right)^{2} \\
=\frac{1}{9}\left[\left(\frac{2 a}{c}+\frac{b}{c}\right)\left(\frac{2 c}{a}+\frac{c}{b}\right)\right]^{2} \\
=\frac{1}{9}\left(4+\frac{2 a}{b}+\frac{2 b}{a}+1\right)^{2} \\
\geqslant \frac{1}{9}(4+2 \times 2+1)^{2}=9,
\end{array}
$$

Equality holds if and only if $a=b=\sqrt{3} c$. Therefore, the minimum value of $\frac{2 a^{2}+b^{2}}{c^{2}}$ is 9.