ta có: \(\frac{a}{\left(a+1\right)\left(b+1\right)}+\frac{b}{\left(b+1\right)\left(c+1\right)}+\frac{c}{\left(c+1\right)\left(a+1\right)}.\)

\(\ge3\sqrt[3]{\frac{a.b.c}{\left(a+1\right)^2.\left(b+1\right)^2.\left(c+1\right)^2}}=\frac{3}{\sqrt[3]{\left(a+1\right)^2.\left(b+1\right)^2.\left(c+1\right)^2}}\)    (vì abc=1)     (*)

Mặt khác: \(\left(a+1\right)^2.\left(b+1\right)^2.\left(c+1\right)^2\ge64abc=64=4^3\)   (vì abc=1)

=> \(\sqrt[3]{\left(a+1\right)^2.\left(b+1\right)^2.\left(c+1\right)^2}\ge4\)   (**)

Từ (*), (**)=> đpcm

Bạn dưới kia làm ngược dấu thì phải,mà bài này hình như là mũ 3

\(\frac{a^3}{\left(a+1\right)\left(b+1\right)}+\frac{a+1}{8}+\frac{b+1}{8}\ge3\sqrt[3]{\frac{a^3\left(a+1\right)\left(b+1\right)}{64\left(a+1\right)\left(b+1\right)}}=\frac{3a}{4}\)

Tương tự rồi cộng lại:

\(RHS+\frac{2\left(a+b+c\right)+6}{8}\ge\frac{3\left(a+b+c\right)}{4}\)

\(\Leftrightarrow RHS\ge\frac{3}{4}\) tại a=b=c=1

Áp dụng bđt bunhiacopxki ta được \(\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\le\left(1+1+1\right)^2=9\)

\(\Leftrightarrow\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\le10\)

bu-nhi a đâu phả vậy đâu bn?

ta có : \(P=\frac{\sqrt{bc}}{a+2\sqrt{bc}}+\frac{\sqrt{ac}}{b+2\sqrt{ac}}+\frac{\sqrt{ab}}{c+2\sqrt{ab}}\le\frac{\frac{1}{2}\left(b+c\right)}{a+b+c}+\frac{\frac{1}{2}\left(a+c\right)}{a+b+c}+\frac{\frac{1}{2}\left(a+b\right)}{a+b+c}\)

\(\Rightarrow P\le\frac{a+b+c}{a+b+c}=1\)

=> GTLN của P là 1 khi a=b=c

ta có: \(x^2+xy-2012x-2013y-2014=0.\)

\(\Leftrightarrow x\left(x+y\right)-2013\left(x+y\right)+x-2013=1\)

\(\Leftrightarrow\left(x+y\right)\left(x-2013\right)+x-2013=1\)

\(\Leftrightarrow\left(x-2013\right)\left(x+y+1\right)=1\)

mà x,y là các số nguyên nên

\(\orbr{\begin{cases}\hept{\begin{cases}x-2013=1\\x+y+1=1\end{cases}}\\\hept{\begin{cases}x-2013=-1\\x+y+1=-1\end{cases}}\end{cases}\Rightarrow\orbr{\begin{cases}\hept{\begin{cases}x=2014\\y=-2014\end{cases}}\\\hept{\begin{cases}x=2012\\y=-2012\end{cases}}\end{cases}}}\)

vậy (x;y)={ (2014;-2014) ;(2012;-2012)}

\(x^2+xy-2012x-2013y-2014=0\) \(0\)

\(\Leftrightarrow x\left(x+y\right)-2013x-2013y+x-2013-1=0\)

\(\Leftrightarrow x\left(x+y\right)-2013\left(x+y\right)+\left(x-2013\right)=1\)

\(\Leftrightarrow\left(x+y\right).\left(x-2013\right)+\left(x-2013\right)=1\)

\(\Leftrightarrow\left(x-2013\right).\left(x+y+1\right)=1\)

Mà x,y lại là số nguyên 

Vậy \(\hept{\begin{cases}\left(x;y\right)=\left(2014;2014\right)\\\left(x;y\right)=\left(2012;2012\right)\end{cases}}\)