chứng minh bđt

N

noinhobinhyen

$(a+b+c)^2 \leq 3(a^2+b^2+c^2) = 9$

$\Rightarrow -3 \leq a+b+c \leq 3$

$\Rightarrow (\dfrac{-3}{3})^3 \leq abc \leq (\dfrac{3}{3})^3$

$\Leftrightarrow -1 \leq abc \leq 1$

Có $|ab|+|bc|+|ca| \leq a^2+b^2+c^2 = 3$

$\Rightarrow |ab|+|bc|+|ca| - abc \leq 3-(-1) = 4$

Dấu [=] xảy ra khi (a;b;c) là hoán vị 3 số (1;1;-1)
 
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