柯西不等式(二维形式)

$$ (a_1^2+a_2^2)(b_1^2+b_2^2) \ge (a_1b_1+a_2b_2)^2 $$

$$ 变形:ac+bd\le \sqrt{(a^2+b^2)(c^2+d^2)} $$

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柯西不等式(积分形式)

$$ [\int_a^bf(x)g(x) dx]^2 \le \int_a^b f^2(x)dx \cdot \int_a^b g^2(x)dx $$

数二真题-2018 - T19