Ta có:
\(\left\{{}\begin{matrix}a^5+a\ge2a^3\\b^5+b\ge2b^3\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a^5\ge2a^3-a\\b^5\ge2b^3-b\end{matrix}\right.\)
\(\Rightarrow a^5+b^5\ge2a^3+2b^3-a-b\)
\(\Rightarrow a^3+b^3\ge2a^3+2b^3-a-b\)
\(\Rightarrow a^3+b^3\le a+b\)
\(\left(a+b\right)\left(a^2-ab+b^2\right)\le a+b\)
\(\Rightarrow a^2+b^2\le1+ab\)
Dấu = xảy ra khi \(a=b=1\)
Đúng 0
Bình luận (0)