Giải thích các bước giải:
Ta có:
$(x_1a-y_1b)^{2n}\ge 0$
$(x_2a-y_2b)^{2n}\ge 0$
$(x_3a-y_3b)^{2n}\ge 0$
$\cdots$
$(x_ma-y_mb)^{2n}\ge 0$
$\to (x_1a-y_1b)^{2n}+(x_2a-y_2b)^{2n}+...+(x_ma-y_mb)^{2n}\ge 0$
Mà $(x_1a-y_1b)^{2n}+(x_2a-y_2b)^{2n}+...+(x_ma-y_mb)^{2n}\le 0$
$\to (x_1a-y_1b)^{2n}+(x_2a-y_2b)^{2n}+...+(x_ma-y_mb)^{2n}=0$
$\to (x_1a-y_1b)^{2n}=(x_2a-y_2b)^{2n}=...=(x_ma-y_mb)^{2n}=0$
$\to x_1a-y_1b=x_2a-y_2b=...=x_ma-y_mb=0$
$\to x_1a=y_1b, x_2a=y_2b, ..., x_ma=y_mb$
$\to \dfrac{x_1}{y_1}=\dfrac{b}{a},\dfrac{x_2}{y_2}=\dfrac{b}{a},...,\dfrac{x_m}{y_m}=\dfrac{b}{a}$
$\to \dfrac{x_1}{y_1}=\dfrac{x_2}{y_2}=...=\dfrac{x_m}{y_m}=\dfrac{b}{a}$
$\to \dfrac{b}{a}=\dfrac{x_1+x_2+...+x_m}{y_1+y_2+...+y_m}$
