giả sử \(a\ge b\ge c\ge0\)
Ta có: \(a+\frac{b}{2}-\frac{a^2+ab+b^2}{a+b}=\frac{1}{2}\left(ab-b^2\right)\ge0\Rightarrow a+\frac{b}{2}\ge\frac{a^2+ab+b^2}{a+b}\)
\(b+\frac{a}{2}-\frac{a^2+ab+b^2}{a+b}=\frac{1}{2}\left(ab-a^2\right)\le0\Rightarrow b+\frac{a}{2}\le\frac{a^2+ab+b^2}{a+b}\)
Tương tự: \(b+\frac{c}{2}\ge\frac{b^2+bc+c^2}{b+c}\ge c+\frac{b}{2};a+\frac{c}{2}\ge\frac{a^2+ac+c^2}{a+c}\ge c+\frac{a}{2}\)
Lại có:+) \(\frac{a^3-b^3}{a+b}+\frac{b^3-c^3}{b+c}+\frac{c^3-a^3}{c+a}\)
\(=\left(a-b\right)\frac{a^2+ab+b^2}{a+b}+\left(b-c\right)\frac{b^2+bc+c^2}{b+c}-\left(a-c\right)\frac{a^2+ac+c^2}{a+c}\)
\(\ge\left(a-b\right)\left(b+\frac{a}{2}\right)+\left(b-c\right)\left(c+\frac{a}{2}\right)-\left(a-c\right)\left(a+\frac{c}{2}\right)\)
\(\ge\frac{-1}{4}\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]\left(1\right)\)
+) \(\frac{a^3-b^3}{a+b}+\frac{b^3-c^3}{b+c}+\frac{c^3-a^3}{c+a}\)
\(=\left(a-b\right)\frac{a^2+ab+b^2}{a+b}+\left(b-c\right)\frac{b^2+bc+c^2}{b+c}-\left(a-c\right)\frac{a^2+ac+c^2}{a+c}\)
\(\le\left(a-b\right)\left(a+\frac{b}{2}\right)+\left(b-c\right)\left(b+\frac{c}{2}\right)-\left(a-c\right)\left(c+\frac{a}{2}\right)\)
\(\le\frac{1}{4}\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]\left(2\right)\)
Từ 1,2 => đpcm
BĐT đã cho tuong duong voi:
\(\left|\frac{\left(a-b\right)\left(b-c\right)\left(c-a\right)\left(ab+bc+ca\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\right|\le\frac{1}{4}\left[\Sigma\left(a-b\right)^2\right]\)
Theo AM-GM ta có: \(\left(ab+bc+ca\right)\le\frac{9}{8}\cdot\frac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{a+b+c}\)
Có: \(VT\le\frac{9}{8}\left|\frac{\sqrt{\left(a-b\right)^2\left(b-c\right)^2\left(c-a\right)^2}}{\left(a+b+c\right)}\right|=\frac{9\sqrt{\left(a-b\right)^2\left(b-c\right)^2\left(c-a\right)^2}}{8\left(a+b+c\right)}\)
Cần chứng minh: \(4\left(a+b+c\right)^2\left[\Sigma\left(a-b\right)^2\right]^2\ge9\left(a-b\right)^2\left(b-c\right)^2\left(c-a\right)^2\)
Rõ ràng \(\Sigma\left(a-b\right)^2\ge3\sqrt[3]{\left(a-b\right)^2\left(b-c\right)^2\left(c-a\right)^2}\)
Cần cm: \(36\left(a+b+c\right)^2\sqrt[3]{\left(a-b\right)^4\left(b-c\right)^4\left(c-a\right)^4}\ge9\sqrt[3]{\left(a-b\right)^6\left(b-c\right)^6\left(c-a\right)^6}\)
Hay \(4\left(a+b+c\right)^2\ge\sqrt[3]{\left(a-b\right)^2\left(b-c\right)^2\left(c-a\right)^2}\)
Tiếp tục là điều hiển nhiên do \(VT\ge4\left[\left(a+b+c\right)^2-3\left(ab+bc+ca\right)\right]\)
\(=2\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]\)
\(\ge6\sqrt[3]{\left(a-b\right)^2\left(b-c\right)^2\left(c-a\right)^2}\ge VP\)
Đẳng thức xảy ra khi \(\hept{\begin{cases}\left(a-b\right)\left(b-c\right)\left(c-a\right)=0\\a-b=b-c=c-a\\a=b=c\end{cases}}\Leftrightarrow a=b=c.\)
