Angle between subspaces of inner product space Let $E$ and $F$ be two non-trivial subspaces of inner product space $H$. Define the angle $\theta$ between $E$ and $F$ as follows
$$\cos\theta=\sup\left\{\frac{|\langle x,y\rangle|}{\|x\|\|y\|}\colon x\in E,y\in F\right\},\quad\theta\in\left[0,\frac{\pi}{2}\right].$$
If there exists a constant $c>0$ such that $\|x+y\|^2\geq c(\|x\|^2+\|y\|^2)$ for all $x\in E$ and $y\in F$, I want to prove that $\theta>0$.
According to the condition, I have
$$\mathrm{Re}\langle x,y\rangle\geq(c-1)\|x\|\|y\|,\quad\forall x\in E,y\in F.$$
But how to use this to obtain the final result?
 A: (Let me work in real inner product spaces for simplicity, otherwise just take $\mathrm{Re}\langle x,y\rangle$ instead of $\langle x,y\rangle)$.
From the existence of $c>0$ such that $\|x+y\|^2\geq c(\|x\|^2+\|y\|^2)$ for all $x\in E$ and $y\in F$, we can write that:
$$\langle x,y\rangle\geq \frac{(c-1)}{2}(\|x\|^2 + \|y\|^2),\quad\forall x\in E,y\in F.$$
now note that in general $A^2 + B^2 \ge 2 AB$, so that you can write:
$$\langle x,y\rangle\geq \frac{(c-1)}{2}(\|x\| + \|y\|) \ge (c-1)\|x\|\|y\|,\quad\forall x\in E,y\in F.$$
Now notice we got $$\frac{\langle x,y\rangle}{\|x\|\|y\|} \ge (c-1)$$
By definition you are dealing with $\theta \in [0, \frac{\pi}{2}]$. To prove that $\theta > 0$ amounts to prove that $\theta \neq 0$, which is equivalent to $\cos(\theta) < 1$, that by definition means $$\sup\left\{\frac{|\langle x,y\rangle|}{\|x\|\|y\|}\colon x\in E,y\in F\right\} <1$$ that is for every $x \in E$ and $y \in F$ we have $$ |\langle x,y\rangle|<\|x\|\|y\| $$
Put them together and get for any $x,y$:
$$\|x\|\|y\| >|\langle x,y\rangle|\ge (c-1)\|x\|\|y\|$$
which is giving you $c < 2$. I would see the contradiction if the condition is "suppose there exists $c >2$", otherwise I do not see anything wrong with $\theta =0$.
A: Here is a solution:
Since $\cos\theta=\sup\{|\langle x,y\rangle|\colon\|x\|=\|y\|=1,x\in E,y\in F\}$, what we need to prove is that
$$\sup\{|\langle x,y\rangle|\colon \|x\|=\|y\|=1,x\in E,y\in F\}<1.$$
According to the condition, I have for $\forall x\in E,y\in F$ and $\|x\|=\|y\|=1$ that
$$\mathrm{Re}\langle x,y\rangle\geq c-1.$$
WLOG, we can assume that $0<c<1$. If $\|x\|=1$, then $\|-x\|=1$, so
$$\mathrm{Re}\langle x,y\rangle=-\mathrm{Re}\langle -x,y\rangle\leq 1-c.$$
Let $z$ be a complex number and define $\mathrm{sgn}z=\frac{\bar{z}}{|z|}$, it’s obvious that $|\mathrm{sgn}z|=1$ and $z\cdot\mathrm{sgn}z=|z|$. Therefore, for any $x\in E,y\in F$ and $\|x\|=\|y\|=1$,
$$\begin{align}
    |\langle x,y\rangle|
& =\mathrm{sgn}\langle x,y\rangle\cdot\langle x,y\rangle \\
& =\bigl\langle \mathrm{sgn}\langle x,y\rangle\cdot x,y\bigr\rangle \\
& =\mathrm{Re} \bigl\langle \mathrm{sgn}\langle x,y\rangle\cdot x,y\bigr\rangle \\
& \leq 1-c
\end{align}$$
since $\|\mathrm{sgn}\langle x,y\rangle\cdot x\|=\|y\|=1$. So finally we get
$$\sup\{|\langle x,y\rangle|\colon \|x\|=\|y\|=1,x\in E,y\in F\}\leq 1-c<1.$$
