Question

[Vũ Như Quỳnh]

giúp mình với!! thanks nha^^

cho a, b, c > 0 thỏa mãn abc=1. cmr:\(\dfrac{a^3}{b\left(c+1\right)}+\dfrac{b^3}{c\left(a+1\right)}+\dfrac{c^3}{a\left(b+1\right)}\ge\dfrac{3}{2}\)

Answer 1

[Lightning Farron]

Áp dụng BĐT AM-GM ta có:

\(\dfrac{a^3}{b\left(c+1\right)}+\dfrac{c+1}{4}+\dfrac{b}{2}\ge3\sqrt[3]{\dfrac{a^3}{b\left(c+1\right)}\cdot\dfrac{c+1}{4}\cdot\dfrac{b}{2}}\)

\(=3\sqrt[3]{\dfrac{a^3}{4\cdot2}\cdot\dfrac{c+1}{c+1}\cdot\dfrac{b}{b}}=3\sqrt[3]{\dfrac{a^3}{8}}=\dfrac{3a}{2}\)

Tương tự cho 2 BĐT còn lại ta cũng có:

\(\dfrac{b^3}{c\left(a+1\right)}\ge\dfrac{3b}{2};\dfrac{c^3}{a\left(b+1\right)}\ge\dfrac{3c}{2}\)

Cộng theo vế 3 BĐT trên ta có:

\(VT+\dfrac{a+b+c+3}{4}+\dfrac{a+b+c}{2}\ge\dfrac{3a+3b+3c}{2}\)

\(\Leftrightarrow VT+\dfrac{3\left(a+b+c\right)}{4}+\dfrac{3}{4}\ge\dfrac{3\left(a+b+c\right)}{2}\)

\(\Leftrightarrow VT+\dfrac{3}{4}\ge\dfrac{3\left(a+b+c\right)}{4}\). Mà theo AM-GM ta có:

\(a+b+c\ge3\sqrt[3]{abc}=3\)\(\Rightarrow VT+\dfrac{3}{4}\ge\dfrac{9}{4}\Rightarrow VT\ge\dfrac{3}{2}=VP\)

Đẳng thức xảy ra khi \(a=b=c=1\)