Question

cho \(a^3+b^3=2\) chứng minh \(0< a+b\le2\)

Answer 1

\(a^3+b^3=2\)

\(\Leftrightarrow\left(a+b\right)^3-3ab\left(a+b\right)=2\)

\(\Rightarrow2=\left(a+b\right)^3-3ab\left(a+b\right)\ge\left(a+b\right)^3-\frac{3}{4}\left(a+b\right)^3=\frac{1}{4}\left(a+b\right)^2\)

\(\Rightarrow\left(a+b\right)^3\le8\)

\(\Rightarrow a+b\le2\)

Answer 2

Vì \(a^3+b^3=2>0\Rightarrow a^3>-b^3\Rightarrow a>-b\Rightarrow a+b>0\)