11. Let $f_{1}(x)=\sqrt{x^{2}+32}, f_{n+1}(x)=\sqrt{x^{2}+\frac{16}{3} f_{n}(x)}, n=1,2, \cdots$. For each $n$, find the real solutions to $f_{n}(x)=3 x$.
$$
f_{1}(x)=\sqrt{x^{2}+32}=3 x \Rightarrow 8 x^{2}=32 \Rightarrow x=2 \text {. }
$$

We will prove by mathematical induction that: $f_{n}(2)=6$. When $n=1$, it is obviously true;
Assume that when $n=k$, the conclusion holds, i.e., $f_{k}(2)=6$.
When $n=k+1$, $f_{k+1}(2)=\sqrt{4+\frac{16}{3} f_{k}(2)}=6$, so the conclusion also holds for $n=k+1$. In summary, for any $n \in \mathbf{N}_{+}$, we have $f_{n}(2)=6$.
Since $f_{n}(x)$ is monotonically increasing on $[0,+\infty)$, the real solution to $f_{n}(x)=3 x$ is only $x=2$.