Biết $a,b,c$ là $3$ cạnh $1$ tam giác. Chứng minh:
$\left|\frac{a}{b}+\frac{b}{c}+\frac{c}{a}-\frac{a}{c}-\frac{c}{b}-\frac{b}{a}\right|<1.$
anh đây e !!!!!!@@ –  Đức Anh 13-03-16 11:03 PM
  | a/b + b/c + c/a - a/c - c/b - b/a | < 1 

Ta có:  | aac/abc + abb/abc + bcc/abc - aab/abc - acc/abc - bbc/abc | < 1 

=> | ( a²c + ab² + bc² - a²b - ac² - b²c) / abc | < 1 

=> | ( a²c + ab² - a²b + bc² - ac² - b²c) / abc | < 1 

=> | [ (a²c + ab² - a²b - abc) + (abc + bc² - ac² - b²c) ] / abc | < 1 

=> | [ a(ac + b² - ab - bc) + c( ab + bc - ac - b²) ] / abc | < 1       (đặt a và c là nhân tử chung )

=> | { a[a(c - b) + b(b - c)] + c[a(b - c) - b(b - c)] } / abc | < 1       (đặt a và b là nhân tử chung )

=> | [ a(b - c)(b - a) + c[(b - c)(a - b)] / abc | < 1 

=> | (a - b)(b - c)(c - a) / abc | < 1 
Mà a, b, c là 3 cạnh tam giác nên: 

- b < (a - b) < a 
-c < (b - c) < b 
-a < (c - a) < c 

=> -abc < (a - b)(b - c)(c - a) < abc 

=> | (a - b)(b - c)(c - a) / abc | < 1 ( Đúng ) ( ĐPCM)
CÓ J SAI SÓT MONG MN BỎ QUA ^^                                                                    jinlion
sai chỗ lào vại –  jinlion2052001 19-03-16 10:13 AM
sai rồi bạn ơi :3 –  tran85295 14-03-16 05:29 PM
Vào đây tham khảo xem đúng không:
https://vn.answers.yahoo.com/question/index?qid=20100516185411AADSeoL

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