+Chứng minh \(a^3+b^3+c^3\ge\frac{\left(a+b+c\right)^3}{9}\text{ }\left(1\right)\)
\(\left(1\right)\Leftrightarrow9\left(a^3+b^3+c^3\right)\ge\left(a+b+c\right)^3\)
\(\Leftrightarrow8\left(a^3+b^3+c^3\right)\ge3ab\left(a+b\right)+3bc\left(b+c\right)+3ca\left(c+a\right)+6abc\)
Ta có: \(a^3+b^3-ab\left(a+b\right)=\left(a-b\right)^2\left(a+b\right)\ge0\Rightarrow ab\left(a+b\right)\le a^3+b^3\), tương tự 2 cụm còn lại.
Theo BĐT Côsi: \(3abc\le a^3+b^3+c^3\)
Cộng theo vế ta có đpcm.
+Chứng minh: \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{9}{a+b+c}\)
Theo BĐT Côsi: \(\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\left(a+b+c\right)\ge3\sqrt[3]{\frac{1}{a}.\frac{1}{b}.\frac{1}{c}}.3\sqrt[3]{a.b.c}=9\)
\(\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{9}{a+b+c}\)
\(\Rightarrow\left(a^3+b^3+c^3\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge\frac{\left(a+b+c\right)^3}{9}.\frac{9}{a+b+c}=\left(a+b+c\right)^2\)
Dấu bằng xảy ra khi 3 biến bằng nhau.
