11. Real numbers $x, y \in(1,+\infty)$, and $x y-2 x-y+1=0$, find the minimum value of $\frac{3}{2} x^{2}+y^{2}$.


$$
y=\frac{2 x-1}{x-1}=2+\frac{1}{x-1} \Rightarrow \frac{3}{2} x^{2}+y^{2}=\frac{3}{2} x^{2}+\frac{1}{(x-1)^{2}}+\frac{4}{x-1}+4,
$$

Let $f(x)=\frac{3}{2} x^{2}+\frac{1}{(x-1)^{2}}+\frac{4}{x-1}+4, f^{\prime}(x)=3 x-\frac{2}{(x-1)^{3}}-\frac{4}{(x-1)^{2}}$, since $f^{\prime}(x)$ is monotonically increasing on $(1,+\infty)$, and $f^{\prime}(x)=0 \Rightarrow 3 x=\frac{2+4(x-1)}{(x-1)^{3}} \Rightarrow 3 x(x-1)^{3}=4 x-2$ $\Rightarrow(x-2)\left(3 x^{3}-3 x^{2}+3 x-1\right)=0 \Rightarrow x=2$, so $f(x)_{\min }=f(2)=6+1+8=15$.