Question

Cho \(x,y,z\ge1\)
Chứng minh \(\dfrac{1}{1+x^3}+\dfrac{1}{1+y^3}+\dfrac{1}{1+z^3}\ge\dfrac{3}{1+xyz}\)

Answer 1

Áp dụng bất đẳng thức Cauchy

\(\Rightarrow VT\ge3\sqrt[3]{\dfrac{1}{\left(1+x^3\right)\left(1+y^3\right)\left(1+z^3\right)}}=\dfrac{3}{\sqrt[3]{\left(1+x^3\right)\left(1+y^3\right)\left(1+z^3\right)}}\)

Chứng minh rằng \(\dfrac{3}{\sqrt[3]{\left(1+x^3\right)\left(1+y^3\right)\left(1+z^3\right)}}\ge\dfrac{3}{1+xyz}\)

\(\Leftrightarrow\left(1+x^3\right)\left(1+y^3\right)\left(1+z^3\right)\le\left(1+xyz\right)^3\)

Áp dụng bất đẳng thức Holder

\(\Rightarrow\left(1+x^3\right)\left(1+y^3\right)\left(1+z^3\right)\ge\left(1+xyz\right)^3\left(đpcm\right)\)

Dấu " = " xảy ra khi \(a=b=c=1\)

Answer 2

Use That : \(\dfrac{1}{a^2+1}+\dfrac{1}{b^2+1}\ge\dfrac{2}{1+ab};\forall a,b\ge1\)