đại 8 : khó quá

H

hoailinhminho

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C

cry_with_me



Theo đầu bài $a^2 + b^2 + c^2 = 1$ nên 1 \geq a,b,c \geq -1

=> ( 1+a)(1+b)(1+c) \geq 0

<=>1+ a + b + c + ab + ac + bc + abc \geq 0 (1)

mà $(1+a+b+c)^2$ \geq 0

<=>2 + 2(a+ b+c) + 2(ab + ac + cb) \geq 0

<=>1 + (a+ b+c) + (ab + ac + cb)\geq 0 (2)

cộg 1,2 => đpcm
 
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