Prove that $\dim(U_1 \cap U_2 \cap U_3) \geq \dim(U_1) + \dim(U_2) + \dim(U_3) − 2n$

Good evening,

Let $U_1$, $U_2$ and $U_3$ be subspaces of the $n$-dimensional vectorspace $V$.

How can I prove that: $$ \dim(U_1 \cap U_2 \cap U_3) \geq \dim(U_1) + \dim(U_2) + \dim(U_3) − 2n $$

I am a beginner and have been despairing of this proof for two hours.

For me it is clear that $\dim(U_1) + \dim(U_2) + \dim(U_3) \leq 3n$. And $\dim(U_1 \cap U_2 \cap U_3) = n$ if $\dim (U_1) + \dim (U_2) + \dim (U_3) = 3n$.

But how can I prove this elegantly? I am grateful for any approach.

You may know that for any subspaces $U,W \subseteq V$ you have

$$\dim (U + W) = \dim U + \dim W - \dim(U \cap W)$$ and therefore

$$\dim (U\cap W) \ge \dim U + \dim W -n$$ as $\dim (U + W) \le \dim V =n$. Based on that you have

$$\begin{aligned} \dim(U_1 \cap U_2 \cap U_3) &\ge \dim (U_1 \cap U_2) + \dim U_3 - n\\ &\ge \dim U_ 1 + \dim U_2 -n + \dim U_3 - n\\ &\ge \dim U_1 + \dim U_2 + \dim U_3 -2n \end{aligned}$$