chứng minh rằng:
$\sum {\dfrac{a}{{b + c}}} + \dfrac{{4(a + b)(b + c)(c + a)}}{{{a^3} + {b^3} + {c^3}}} \ge 5$
Edited by dcmtltvclh, 19-04-2009 - 18:29.
Edited by dcmtltvclh, 19-04-2009 - 18:29.
Edited by tuan101293, 19-04-2009 - 21:19.
KT-PT
Do unto others as you would have them do unto you.
chứng minh $f(a,b,c) \geq f(a,b,0)$kiểu gì dzậy ta?$f(a,b,c) \geq f(a,b,0)$
đặt $\dfrac{a}{b}=x$
suy ra:
$f(a,b,0)-5=\dfrac{(x^2-3*x+1)^2}{x*(x^2-x+1)}\geq 0$(ĐPCM)
Edited by muctieu-5, 20-04-2009 - 02:08.
KT-PT
Do unto others as you would have them do unto you.
Bài này của Phạm Sinh Tân. Các bạn khi post các bài toán thì nên ghi rõ nguồn gốc (nếu có) nhé.cho $a,b,c \ge 0$
chứng minh rằng:
$\sum {\dfrac{a}{{b + c}}} + \dfrac{{4(a + b)(b + c)(c + a)}}{{{a^3} + {b^3} + {c^3}}} \ge 5$
Edited by tuan101293, 27-04-2009 - 10:42.
KT-PT
Do unto others as you would have them do unto you.
lech. tam dau co' nghia~ la` kok S.O.S dc. (chieu` thi ve` minh` thu? lam` xem sao da~)cực trị lệch tâm mà phân tích SOS à em
Edited by tuan101293, 27-04-2009 - 10:39.
Edited by nguyen_ct, 27-04-2009 - 12:56.
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