Determine the maximum value of a real number $c$ such that for any two unit vectors $\vec{v_1}$ and $\vec{v_2}$ in the 2D plane, there exists a unit vector $\vec{v}$ satisfying $|\vec{v_1} \cdot \vec{v}| \cdot |\vec{v_2} \cdot \vec{v}| \geq c$.
My thoughts:
Since rotation does not affect the result, let $\vec{v_1} = (1, 0), \vec{v_2} = (0, 1)$, and $\vec{v} = (\cos \theta, \sin \theta)$ $\Rightarrow |\vec{v_1} \cdot \vec{v}| \cdot |\vec{v_2} \cdot \vec{v}| = \cos\theta\sin\theta = \dfrac{\sin 2\theta}{2} \le \boxed{\frac12}$
So I guess $c \le \dfrac12$. And I need strict proof.
$$|v_1\cdot v|\cdot |v_2 \cdot v| = |v_1||v||cos(\theta_1)||v_2||v||cos(\theta_2)|$$
Since they are all unit vectors: $$|v_1||v||cos(\theta_1)||v_2||v||cos(\theta_2)| = |cos(\theta_1)cos(\theta_2)|$$
To find the maximum value for $c$: We simply need to find the maximum value of $c$ for which: $$|cos(\theta_1)cos(\theta_2)|\geq c$$
The maximum value for $|cos(\theta_1)cos(\theta_2)|$ is $1$.
This can be seen by making $\theta_1 = 0, \theta_2 = 0$.
Thus the maximum value for which: $$|cos(\theta_1)cos(\theta_2)|\geq c$$
is $c = 1$.
Hence $c = 1$