[Toán 9] Tìm Max

H

huynhbachkhoa23

$P=\sum \dfrac{a^2}{(a+b)(a+c)}=1-\dfrac{2abc}{(a+b)(b+c)(c+a)}$
Ngoài ra $(a+b)(b+c)(c+a)\ge 8abc$ nên $-\dfrac{2abc}{(a+b)(b+c)(c+a)}\ge -\dfrac{1}{4}$
Do đó $P\ge \dfrac{3}{4}$
 
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