toán hsg

V

vipboycodon

Theo bdt cauchy schwarz ta có :
$\dfrac{1}{a^2+b^2+c^2}+\dfrac{2009}{ab+bc+ac}$
= $\dfrac{1}{a^2+b^2+c^2}+\dfrac{1}{ab+bc+ac}+\dfrac{1}{ab+bc+ac}+\dfrac{2007}{ab+bc+ac}$
$\ge \dfrac{9}{(a+b+c)^2}+\dfrac{2007}{\dfrac{(a+b+c)^2}{3}} \ge 670$
Dấu "=" xảy ra khi $a = b = c = 1$
 
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