Предмет: Алгебра, автор: Полииинаа

Докажите: 2х²+2у² ≥ (х+у)²
Заранее спасибо)

Ответы

Автор ответа: LFP
0
2x² + 2y² - (x+y)² ≥ 0
2x² + 2y² - x² - 2xy - y² ≥ 0
x² + y² - 2xy ≥ 0
(x - y)² ≥ 0
это верно всегда (квадрат не может быть отрицательным числом)))
Автор ответа: tseluyko58
0
2х²+2у²-(х+у)²= 2х²+2у²-х²-2ху-у²=х²-2ху+у²=(х-у)²≥0 что и требовалось доказать
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