[Tooán 9] cực trị

H

huynhbachkhoa23

$\sum\limits_{cyc}\dfrac{a}{b+c+2a}\le \dfrac{1}{4}\sum\limits_{cyc}\left( \dfrac{a}{a+b}+\dfrac{a}{a+c}\right) = \dfrac{3}{4}$. Do đó $M\le \sqrt{3\sum\limits_{cyc}\dfrac{a}{b+c+2a}} \le \dfrac{3}{2}$
 
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