Cho $a,b,c$ la 3 canh cua 1 tam giac. Tim Min cua:
 $P=\frac{4a}{b+c-a} + \frac{9b}{c+a-b} + \frac{16c}{a+b-c}$
trời...thôi –  hờ hờ 26-04-16 08:22 PM
mà giải r thì nhăng chả được –  Kira Kira 26-04-16 08:21 PM
k hiểu. à roi –  Kira Kira 26-04-16 08:20 PM
t vừa vote up cho m m bị đứa mô đ mô vote down –  hờ hờ 26-04-16 08:14 PM
lm gì. con chó –  Kira Kira 26-04-16 08:07 PM
vào ô chat! :v –  hờ hờ 26-04-16 08:04 PM
ở mô chả đuoc –  Kira Kira 26-04-16 08:00 PM
móc mô ra đây –  hờ hờ 26-04-16 07:54 PM
http://toan.hoctainha.vn/Hoi-Dap/Cau-Hoi/127171/gtnn
thanks!lam nhanh thật –  Kira Kira 26-04-16 07:42 PM
Đặt b + ca = 2x ; c + a b = 2y ;a + bc = 2z thì x,y,z > 0
Ta có a = y + z ; b = z + x ; c = x + y
Áp dụng BĐT Cauchy ta có
2P= 4((y+z)/x)+9((z+x)/y)+16((x+y)/z)
= (4y/x + 9x/y) + (4z/x+ 16x/z) + (9z/y + 16y/z)12 + 16 + 24 = 52
P 26
Đẳng thức xảy ra khi 3z = 4y = 6x
thanks nha –  Kira Kira 26-04-16 07:40 PM

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