$$ (a_1^2+a_2^2)(b_1^2+b_2^2) \ge (a_1b_1+a_2b_2)^2 $$
$$ 变形:ac+bd\le \sqrt{(a^2+b^2)(c^2+d^2)} $$
$$ [\int_a^bf(x)g(x) dx]^2 \le \int_a^b f^2(x)dx \cdot \int_a^b g^2(x)dx $$
数二真题-2018 - T19