Question 87: Let unit vectors $\vec{a} 、 \vec{b}$ have an acute angle between them. If for any
$$
(x, y) \in\{(x, y)|| x \vec{a}+y \vec{b} \mid=1, x y \geq 0\}
$$

it holds that $|x+2 y| \leq \frac{8}{\sqrt{15}}$, find the minimum possible value of $\vec{a} \cdot \vec{b}$.
Question 87, Solution: Noting that a, b are unit vectors, hence
$$
|x \vec{a}+y \vec{b}|=1 \Leftrightarrow(x \vec{a}+y \vec{b})^{2}=1 \Leftrightarrow x^{2}+y^{2}+2 \vec{a} \cdot \vec{b} \cdot x y=1
$$

Let $\vec{a} \cdot \vec{b}=k$, we need to find the smallest positive real number $k$, such that for any non-negative real numbers $x, y$ satisfying $x^{2}+y^{2}+2 k x y=1$, we have $|x+2 y| \leq \frac{8}{\sqrt{15}}$. In fact,
$$
\begin{array}{l}
|x+2 y| \leq \frac{8}{\sqrt{15}} \\
\Leftrightarrow 15 x^{2}+60 x y+60 y^{2} \leq 64 \\
\Leftrightarrow 15 x^{2}+60 x y+60 y^{2} \leq 64\left(x^{2}+y^{2}+2 k x y\right) \\
\Leftrightarrow 2(30-64 k) x y \leq 49 x^{2}+4 y^{2}
\end{array}
$$

And $\min \left\{49 x^{2}+4 y^{2} \mid x y \geq 0\right\}=2 \times 14 x y$, hence $30-64 k \leq 14 \Rightarrow k \geq \frac{1}{4}$. Therefore, the smallest possible value of $\vec{a} \cdot \vec{b}$ is $\frac{1}{4}$.