đặt \(a+b=x,b+c=y;c+a=z\)
ta có \(\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}=1\Rightarrow3-\frac{1}{x+1}-\frac{1}{y+1}-\frac{1}{z+1}=1\) \(\)
=> \(\frac{x}{x+1}+\frac{y}{y+1}+\frac{z}{z+1}=1\)
=> \(\frac{y}{y+1}+\frac{z}{z+1}=1-\frac{x}{x+1}=\frac{1}{x+1}\)
Áp dụng bđt cô si ta có \(\frac{y}{y+1}+\frac{z}{z+1}\ge2\sqrt{\frac{yz}{\left(y+1\right)\left(z+1\right)}}\)
=> \(\frac{1}{x+1}\ge2\sqrt{\frac{yz}{\left(y+1\right)\left(z+1\right)}}\)
tương tự ta có
\(\frac{1}{y+1}\ge2\sqrt{\frac{zx}{\left(z+1\right)\left(x+1\right)}}\)
\(\frac{1}{z+1}\ge2\sqrt{\frac{xy}{\left(x+1\right)\left(y+1\right)}}\)
nhân từng vế của 3 bđt cùng chièu ta có
\(\frac{1}{x+1}.\frac{1}{y+1}.\frac{1}{z+1}\ge8\sqrt{\frac{x^2y^2z^2}{\left(x+1\right)^2\left(y+1\right)^2\left(z+1\right)^2}}=8.\frac{xyz}{\left(x+1\right)\left(y+1\right)\left(z+1\right)}\)
=> \(1\ge8xyz\Rightarrow xyz\le\frac{1}{8}\Rightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)\le\frac{1}{8}\)
