Toán 9

V

vipboycodon

Bài 2:
$a^2+b^2+c^2+d^2 \ge ab+ac+ad$
<=> $\dfrac{a^2}{4}-ab+b^2+\dfrac{a^2}{4}-ac+c^2+\dfrac{a^2}{4}-ad+d^2+\dfrac{a^2}{4} \ge 0$
<=> $(\dfrac{a}{2}-b)^2+(\dfrac{a}{2}-c)^2+(\dfrac{a}{2}-d)^2+\dfrac{a^2}{4} \ge 0$
Dấu "=" xảy ra khi $a = b = c = d = 0$
 
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