a, Vì a,b,c dương nên : \(a+b+c\ge3\sqrt[3]{abc}\) (1)
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge3\sqrt[3]{\frac{1}{abc}}\) (2)
Nhân vế theo vế 1 và 2 ta có : \(\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge9\sqrt[3]{\frac{abc}{abc}}=9\)
Mà a+b+c=1 nên \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge9\)
\(a+b+c=0\Leftrightarrow a+b=-c\)
\(\Rightarrow\left(a+b\right)^5=-c^5\)
\(\Leftrightarrow a^5+b^5+5a^4b+10a^3b^2+10a^2b^3+5ab^4=-c^2\)
\(\Leftrightarrow a^5+b^5+c^5=-5ab\left(a^3+2a^2b+2ab^2+b^3\right)\)
\(\Leftrightarrow a^5+b^5+c^5=-5ab\left(a+b\right)\left(a^2+ab+b^2\right)\)
\(\Leftrightarrow a^5+b^5+c^5=5abc\left(a^2+ab+b^2\right)\)
\(\Leftrightarrow2\left(a^5+b^5+c^5\right)=5abc\left(2a^2+2ab+2b^2\right)\)
\(\Leftrightarrow2\left(a^5+b^5+c^5\right)=5abc\left(a^2+b^2+\left(a+b\right)^2\right)=5abc\left(a^2+b^2+c^2\right)\)