Theo bđt AM-GM :
\(\frac{a^3}{\left(b+1\right)\left(c+1\right)}+\frac{b+1}{8}+\frac{c+1}{8}\)\(\ge3\sqrt[3]{\frac{a^3}{\left(b+1\right)\left(c+1\right)}\cdot\frac{b+1}{8}\cdot\frac{c+1}{8}}=\frac{3a}{4}\)
Dấu "=" xảy ra \(\Leftrightarrow\frac{a^3}{\left(b+1\right)\left(c+1\right)}=\frac{b+1}{8}=\frac{c+1}{8}\)
\(\Leftrightarrow2a=b+1=c+1\)
+ Tương tự ta cm đc :
\(\frac{b^3}{\left(c+1\right)\left(a+1\right)}+\frac{c+1}{8}+\frac{a+1}{8}\ge\frac{3b}{4}\)
Dấu "=" xảy ra \(\Leftrightarrow2a=b+1=c+1\)
\(\frac{c^3}{\left(a+1\right)\left(b+1\right)}+\frac{a+1}{8}+\frac{c+1}{8}\ge\frac{3c}{4}\)
Dấu "=" xảy ra \(\Leftrightarrow2a=a+1=b+1\)
Do đó : \(\frac{a^3}{\left(b+1\right)\left(c+1\right)}+\frac{b^3}{\left(c+1\right)\left(a+1\right)}+\frac{c^3}{\left(a+1\right)\left(b+1\right)}+\frac{a+b+c+3}{4}\ge\frac{3}{4}\left(a+b+c\right)\)
\(\Rightarrow\frac{a^3}{\left(b+1\right)\left(c+1\right)}+\frac{b^3}{\left(c+1\right)\left(a+1\right)}+\frac{c^3}{\left(a+1\right)\left(b+1\right)}\ge\frac{1}{2}\left(a+b+c\right)-\frac{3}{4}\)
\(\ge\frac{1}{2}\cdot3\sqrt[3]{abc}-\frac{3}{4}=\frac{3}{4}\)
Dấu "=" xảy ra <=> a = b = c = 1
Áp dụng bđt AM-GM
\(\frac{a^3}{\left(1+b\right)\left(1+c\right)}+\frac{1+b}{8}+\frac{1+c}{8}\ge\frac{3}{4}a\)
\(\frac{b^3}{\left(1+c\right)\left(1+a\right)}+\frac{1+c}{8}+\frac{1+b}{8}\ge\frac{3}{4}b\)
\(\frac{c^3}{\left(1+a\right)\left(1+b\right)}+\frac{1+a}{8}+\frac{1+b}{8}\ge\frac{3}{4}c\)
\(\Rightarrow A+\frac{6+2a+2b+2c}{8}\ge\frac{3}{4}\left(a+b+c\right)\)
\(\Rightarrow A+\frac{3}{4}\ge\frac{1}{2}\left(a+b+c\right)\ge\frac{3}{2}\sqrt[3]{abc}=\frac{3}{2}\)
\(\Rightarrow A\ge\frac{3}{4}\)
\("="\Leftrightarrow a=b=c=1\)
