lower bound for vector inner products Suppose $A, B, C$ are three complex valued vectors, and denote $(A, B)= \sum_i A_i^* B_i$ as the inner product. Here $A_i$ is the $i$-th element of $A$. All satisfy normalization condition:
$$(A, A)=1,   (B, B)=1,    (C, C)=1$$

Now suppose $|(A, B)|>1-\epsilon_1$, and $|(B, C)|>1-\epsilon_2$ for some small numbers $\epsilon_1, \epsilon_2>0$. Here $|...|$ is the standard norm.  Then is there a lower bound for 
  $|(A, C)|$
  ?

Intuitively it is so, because $B$ is close to $A$ and $C$, so $A$ and $C$ must be close to each other.   
 A: You can get such a bound. Note that we can assume without loss of generality that the first two inner products are real and positive, i.e.
\begin{gather}
\vert \langle A,B\rangle\vert=Re(\langle A,B\rangle)>1-\epsilon_1\\
\vert \langle B,C\rangle\vert=Re(\langle B,C\rangle)>1-\epsilon_2.
\end{gather}
We can do this by first rotating $A$ by some complex phase to make the first line hold, and then rotating $B$ by some complex phase to make the second line hold.
Then, by the Polarization Identity, we find that 
\begin{align*}
Re(\langle A,C\rangle)&=\frac{1}{2}(\|A\|+\|C\|-\|A-C\|^2)\\
&=1-\frac{\|A-C\|^2}{2}\\
&\geq 1-\frac{(\|A-B\|+\|B-C\|)^2}{2}\\
\end{align*}
Let's now estimate these latter two terms. Again by the Polarization Identity,
\begin{equation}
\|A-B\|=\sqrt{2-2Re(\langle A,B\rangle)}<\sqrt{2\epsilon_1},
\end{equation}
and similarly,
\begin{equation}
\|B-C\|<\sqrt{2\epsilon_2},
\end{equation}
so plugging this in, we obtain
\begin{equation}
\vert \langle A,C\rangle\vert\geq Re(\langle A,C\rangle)>1-\frac{(\sqrt{2\epsilon_1}+\sqrt{2\epsilon_2})^2}{2}=1-(\sqrt{\epsilon_1}+\sqrt{\epsilon_2})^2.
\end{equation}
I don't know how tight this is, but it seems roughly comparable to the best you can hope for with real inner products using angles.
