Question

Cho a,b,c >0 thỏa mãn abc=1.CMR \(\dfrac{a^3}{\left(1+b\right)\left(1+c\right)}+\dfrac{b^3}{\left(1+c\right)\left(1+a\right)}+\dfrac{c^3}{\left(1+a\right)\left(1+b\right)}\ge\dfrac{3}{4}\)

Answer 1

Áp dụng bđt AM-GM ta có :

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

Tương tự cho `2` bđt còn lại ta có :

\(\dfrac{b^3}{\left(1+c\right)\left(1+a\right)}+\dfrac{c+1}{8}+\dfrac{a+1}{8}\ge\dfrac{3b}{4}\\ \dfrac{c^3}{\left(1+a\right)\left(1+b\right)}+\dfrac{a+1}{8}+\dfrac{b+1}{8}\ge\dfrac{3c}{4}\)

Cộng theo vế 3bđt ta có :

\(VT+\dfrac{2\left(a+b+c+3\right)}{8}\ge\dfrac{3\left(a+b+c\right)}{4}\\\Leftrightarrow VT+\dfrac{2\left(3\sqrt[3]{abc}+3\right)}{8}\ge\dfrac{3\cdot3\sqrt[3]{abc}}{4}\\ \Leftrightarrow VT\ge\dfrac{3}{4}=VP \)

khi `a=b=c=1`