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sgwriadaufa24r

sgwriadaufa24r

Answered question

2022-06-04

a ( a + b ) 4 a 2 + a b + b 2 + b ( b + c ) 4 b 2 + b c + c 2 + c ( c + a ) 4 c 2 + c a + a 2 1
I've got stuck at this problem:
Firstly, I've thought this :
a 2 2 a b + b 2 0
4 a 2 + a b + b 2 3 ( a b + a 2 )
a ( a + b ) 4 a 2 + a b + b 2 1 3
Similarly we obtain that
b ( b + c ) 4 b 2 + b c + c 2 1 3
Summing all, we obtain the inequality.
Is this way correct? (I have doubts about my solution because this problem was found in a math magazine which usually has difficult problems(at least for me)). Or is there a another way?
Thanks!

Answer & Explanation

hatasky815jd

hatasky815jd

Beginner2022-06-05Added 4 answers

The inequality is equivalent to:
c y c 3 a ( a + b ) 4 a 2 + a b + b 2 3 0 c y c 1 3 a ( a + b ) 4 a 2 + a b + b 2 = c y c a 2 2 a b + b 2 4 a 2 + a b + b 2 = c y c ( a b ) 2 ( a b ) 2 + 3 a ( a + b ) 0 c y c ( a b ) 2 ( 7 4 a 1 4 b ) 2 + 15 16 ( a + b ) 2
Which is obvious.

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