toán 9 tìm min

T

transformers123

$M=\sum \sqrt{a^2-ab+b^2} = \sum \sqrt{(a-\dfrac{b}{2})^2+\dfrac{3b^2}{4}} \ge \sqrt{[a+b+c-(\dfrac{a+b+c}{2})]^2+[\dfrac{\sqrt{3}}{2}(a+b+c)]^2}$

$\iff M \ge 1$

Dấu "=" xảy ra khi $a=b=c=\dfrac{1}{3}$
 
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