Sửa đề: \(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+\frac{3}{a+b+c}\ge4\)
\(\Leftrightarrow\frac{a^2c+b^2a+c^2b}{abc}+\frac{3}{a+b+c}\ge4\)
\(\Leftrightarrow P=a^2c+b^2a+c^2b+\frac{3}{a+b+c}\ge4\)
Ta có:
\(a^2c+a^2c+b^2a\ge3\sqrt[3]{a^3.\left(abc\right)^2}=3a\)
\(b^2a+b^2a+c^2b\ge3\sqrt[3]{b^3\left(abc\right)^2}=3b\)
\(c^2b+c^2b+a^2c\ge3\sqrt[3]{c^3\left(abc\right)^2}=3c\)
Cộng vế với vế: \(a^2c+b^2a+c^2b\ge a+b+c\)
\(\Rightarrow P\ge a+b+c+\frac{3}{a+b+c}=\frac{a+b+c}{3}+\frac{3}{a+b+c}+\frac{2}{3}\left(a+b+c\right)\)
\(\Rightarrow P\ge2\sqrt{\frac{3\left(a+b+c\right)}{3\left(a+b+c\right)}}+\frac{2}{3}.3\sqrt[3]{abc}=4\)
Dấu "=" xảy ra khi \(a=b=c=1\)
