MOSP $2003$ CMR với mọi số thực dương $a,b,c$, ta luôn có: 
$\frac{1}{a(b+1)}+\frac{1}{b(c+1)}+\frac{1}{c(a+1)}\geq\frac{3}{\sqrt[3]{abc}(1+\sqrt[3]{abc})}$
Ta có:$(1+abc)(\frac{1}{a(b+1)}+\frac{1}{b(c+1)}+\frac{1}{c(a+1)})+3=\Sigma \frac{1+abc+a+ab}{a(1+b)}=\Sigma\frac{1+a}{a(1+b)}+\Sigma \frac{b(c+1)}{1+b}$
Sử dụng 2 lần BĐT AM-GM ta được:
$\Sigma\frac{1+a}{a(1+b)}+\Sigma\frac{b(c+1)}{1+b}\geq\frac{3}{\sqrt[3]{abc}}+3\sqrt[3]{abc}$
Ta phải CM:$\frac{\frac{3}{\sqrt[3]{abc}}+3\sqrt[3]{abc}-3}{1+abc}\geq \frac{3}{\sqrt[3]{abc}(1+\sqrt[3]{abc})}$(1)
Quy đồng ra điều luôn đúng do(1) là 1 đống nhất thức:D
sao k cưng –  Lionel Messi 13-11-16 04:40 AM
giỏi thế –  Lionel Messi 24-08-16 08:07 PM
http://toan.hoctainha.vn/Hoi-Dap/Cau-Hoi/137435/bat-dang-thuc/39363#39363
Đặt $$abc=k^3$$.Khi đó tồn tại các số nguyên dương x,y,z sao cho
$$a=\frac{kx}{y};b=\frac{kz}{x};c=\frac{ky}{z}$$.Ta viết lại Bđt đã cho dưới dạng 
$$\frac{y}{x+kz}+\frac{x}{z+ky}+\frac{z}{y+kx}\ge \frac{3}{1+k}$$
Áp dụng Bđt Cauchy-Schwarz ta có: 
$$VT\ge \frac{\left(x+y+z\right)^2}{\left(1+k\right)\left(xy+yz+xz\right)}\ge \frac{3}{1+k}=VP$$
P/s:ngoài ra có thể dùng Holder nếu bn cần thì nhắn cho mình

anh học sinh lớp mấy vậy ạ ? –  Hàn Thiên Dii 12-11-16 07:54 AM

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