Question

cho a;b;c thoă mãn là 3 số dương và abc=1

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

Answer 1

Áp dụng BĐT AM-GM ta có:

\(\dfrac{1}{a^3\left(b+c\right)}+\dfrac{a\left(b+c\right)}{4}\ge2\sqrt{\dfrac{1}{a^3\left(b+c\right)}.\dfrac{a\left(b+c\right)}{4}=2\sqrt{\dfrac{1}{4a^2}=\dfrac{1}{a}=\dfrac{abc}{a}=bc}}\)

Tương tự:

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

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

Cộng theo vế:

\(\Rightarrow VT+\dfrac{ab+bc+ac}{2}\ge ab+bc+ac\)

\(\Rightarrow VT\ge\dfrac{ab+bc+ac}{2}\)

Tiếp tục áp dụng AM-GM: \(ab+bc+ac\ge3^3\sqrt{a^2b^2c^2}=3\)

\(\Rightarrow VT\ge\dfrac{3}{2}\left(đpcm\right)\)

Dấu bằng xảy ra khi a=b=c=1