C/m:a=b=c=0

H

huynhbachkhoa23

Theo phương trình:
$a^2+b^2+c^2=\dfrac{a^2-b^2}{c^2+5}+\dfrac{b^2-c^2}{a^2+3}+\dfrac{c^2-a^2}{b^2+4} \le \dfrac{a^2}{c^2+5}+\dfrac{b^2}{a^2+3}+\dfrac{c^2}{b^2+4} \le \dfrac{a^2}{5}+\dfrac{b^2}{3}+\dfrac{c^2}{4} \;\; (1)$

Nhưng $a^2 \ge \dfrac{1}{5}a^2; b^2 \ge \dfrac{1}{3}b^2; c^2 \ge \dfrac{1}{4}c^2$

Vì vậy mà $\dfrac{a^2}{5}+\dfrac{b^2}{3}+\dfrac{c^2}{4} \le a^2+b^2+c^2\;\;(2)$

Từ $(1), (2)$ ta có $a=b=c=0$
 
Last edited by a moderator:
You must log in or register to reply here.
Top Bottom