Đặt \(\left(a+b;b+c;c+a\right)=\left(x;y;z\right)\)
\(\Rightarrow\frac{1}{1+x}+\frac{1}{1+y}+\frac{1}{1+z}=2\)
\(\Rightarrow\frac{1}{1+x}=1-\frac{1}{1+y}+1-\frac{1}{1+z}=\frac{y}{1+y}+\frac{z}{1+z}\ge2\sqrt{\frac{yz}{\left(1+y\right)\left(1+z\right)}}\)
Tương tự: \(\frac{1}{1+y}\ge2\sqrt{\frac{zx}{\left(1+z\right)\left(1+x\right)}}\) ; \(\frac{1}{1+z}\ge2\sqrt{\frac{xy}{\left(1+x\right)\left(1+y\right)}}\)
Nhân vế với vế:
\(\frac{1}{\left(1+x\right)\left(1+y\right)\left(1+z\right)}\ge\frac{8xyz}{\left(1+x\right)\left(1+y\right)\left(1+z\right)}\)
\(\Rightarrow8xyz\le1\Rightarrow A=xyz\le\frac{1}{8}\)
\(\Rightarrow A_{max}=\frac{1}{8}\) khi \(x=y=z=\frac{1}{2}\) hay \(a=b=c=\frac{1}{4}\)
Ta có :
\(\frac{1}{a+b+1}=\left(1-\frac{1}{b+c+1}\right)+\left(1-\frac{1}{a+c+1}\right)=\frac{b+c}{b+c+1}+\frac{a+c}{a+c+1}\)
\(\ge2\sqrt{\frac{\left(b+c\right)\left(a+c\right)}{\left(b+c+1\right)\left(a+c+1\right)}}\)
Tương tự ta cũng có :
\(\frac{1}{b+c+1}\ge2\sqrt{\frac{\left(a+b\right)\left(a+c\right)}{\left(a+b+1\right)\left(a+c+1\right)}}\)
\(\frac{1}{a+c+2}\ge2\sqrt{\frac{\left(a+b\right)\left(b+c\right)}{\left(a+b+1\right)\left(b+c+1\right)}}\)
Nhân 3 BĐT trên ta được :
\(\frac{1}{\left(a+b+1\right)\left(b+c+1\right)\left(a+c+1\right)}\ge\frac{8\left(a+b\right)\left(b+c\right)\left(a+c\right)}{\left(a+b+1\right)\left(b+c+1\right)+\left(a+c+1\right)}\)
\(\Rightarrow\left(a+b\right)\left(b+c\right)\left(a+c\right)\le\frac{1}{8}\)
\(Max_A=\frac{1}{8}\) khi \(a=b=c=\frac{1}{4}\)
