tìm gtnh

H

hieunguyenhoang1

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B

braga

Xét điều kiện: $a^3c+b^3a+c^3b=abc\iff \dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{a}=1$
$1=\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{a}\ge \dfrac{(a+b+c)^2}{a+b+c}=a+b+c\implies a+b+c\le 1$
Ta có:
$$P=\sum\dfrac{b}{a^2+ab}=\sum\dfrac{1}{\dfrac{a^2}{b}+a}\ge \dfrac{9}{\sum \dfrac{a^2}{b}+\sum a}\ge \dfrac{9}{2}$$
 
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