Giả sử \(x\le0;a,b\ge0\)
Ta có \(c=-a-b\) và hàm \(f\left(x\right)\) lẻ nên
\(f\left(a\right).f\left(b\right)+f\left(b\right).f\left(c\right)+f\left(c\right).f\left(a\right)\le0\)
\(\Leftrightarrow f\left(a\right).f\left(b\right)\le-f\left(b\right).f\left(c\right)-f\left(c\right).f\left(a\right)=-f\left(c\right)\left[f\left(a\right)+f\left(b\right)\right]\)
\(\Leftrightarrow f\left(a\right).f\left(b\right)\le f\left(-c\right)\left[f\left(a\right)+f\left(b\right)\right]\)
\(\Leftrightarrow f\left(a\right).f\left(b\right)\le f\left(a+b\right).f\left(a\right)+f\left(a+b\right).f\left(b\right)\left(1\right)\)
Do \(f\left(x\right)\) đồng biến trên \(R\) nên
\(\left\{{}\begin{matrix}a+b\ge a\\a+b\ge b\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}f\left(a+b\right)\ge f\left(a\right)\\f\left(a+b\right)\ge f\left(b\right)\end{matrix}\right.\)
\(f\left(a+b\right).f\left(a\right)+f\left(a+b\right).f\left(b\right)\ge\left[f\left(a\right)\right]^2+\left[f\left(b\right)\right]^2\ge2f\left(a\right)f\left(b\right)\ge f\left(a\right)f\left(b\right)\)
\(\Rightarrow\left(1\right)\text{ đúng }\left(\text{đpcm}\right)\)
