$$\frac{2}{\frac{1}{a}+\frac{1}{b}}\leq\sqrt{ab}\leq\frac{a+b}{2}\leq\sqrt{\frac{a^2+b^2}{2}},$$

$$H_2\leq G_2\leq A_2\leq Q_2,$$

$$P^{n}_2=\left(\frac{a^n+b^n}{2}\right)^{1/n}$$

$$H_2=P_2^{-1}, A_2=P_2^{1}, Q_2=P_2^{2}.$$

$$G_2=\lim_{n\to0}P^{n}_2:=P_2^{0}.$$

$$P^{-1}_2\leq P^{0}_2\leq P^{1}_2 \leq P^{2}_2.$$

$$\forall n<m\in\mathbb R, \\ P_2^{n}=\left(\frac{a^n+b^n}{2}\right)^{1/n}\leq \left(\frac{a^m+b^m}{2}\right)^{1/m}=P_2^{m}.$$

A. $\frac{1}{ab}\leq\frac{1}{4}$ B. $\frac{1}{a}+\frac{1}{b}\leq1$ C. $\sqrt{ab}\geq 4$ D. $a^2+b^2\geq 8$

$$\frac{1}{y/2}+\frac{1}{x/8}=1,$$

$$2=\frac{2}{\frac{1}{y/2}+\frac{1}{x/8}}\leq \sqrt{\frac{y}{2}\cdot\frac{x}{8}}=\frac{\sqrt{xy}}{4}$$

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2020年11月30日