Bất Đẳng Thức Cosi

H

hien_vuthithanh

Ta có:

$ab^2+bc^2+ca^2 \le ab^2+bc^2+ca^2+abc=a(b+c)^2+c(c-a)(b-a)$

Giả sử $b \le a \le c$ thì $ab^2+bc^2+ca^2 \le a(b+c)^2 $

Ta có : $a(b+c)^2 =\dfrac{1}{2}.2a.(b+c)(b+c) \le \dfrac{1}{2}( \dfrac{2a+2b+2c}{3})^3=\dfrac{1}{2}(\dfrac{2.3}{3})^3=4$

$\rightarrow$ đpcm
 
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