Ta có
\(\left(a^2+b^2+c^2\right)\left(a^3+b^3+c^3\right)=a^5+a^2b^3+a^2c^3+a^3b^2+b^5+b^2c^3+a^3c^2+b^3c^2+c^5\)
\(\Rightarrow a^5+b^5+c^5=\left(a^2+b^2+c^2\right)\left(a^3+b^3+c^3\right)-a^2b^2\left(a+b\right)-b^2c^2\left(b+c\right)-a^2c^2\left(a+c\right)\)
Do a+b+c=0
=> a+b=-c; b+c=-a; a+c=-b
\(\Rightarrow a^5+b^5+c^5=\left(a^2+b^2+c^2\right)\left(a^3+b^3+c^3\right)+a^2b^2c+ab^2c^2+a^2bc^2=\)
\(=\left(a^2+b^2+c^2\right)\left(a^3+b^3+c^3\right)+abc\left(ab+bc+ac\right)=\)
\(=\left(a^2+b^2+c^2\right)\left[\left(a+b\right)^3-3ab\left(a+b\right)+c^3\right]+abc\left(ab+bc+ac\right)=\)
\(=\left(a^2+b^2+c^2\right).\left[\left(-c^3\right)-3ab.\left(-c\right)+c^3\right]+abc\left(ab+bc+ac\right)=\)
\(=\left(a^2+b^2+c^2\right).3abc+abc\left(ab+bc+ab\right)=\)
\(=abc.\left[3\left(a^2+b^2+c^2\right)+ab+bc+ac\right]=\)
\(=abc\left[\dfrac{5}{2}.\left(a^2+b^2+c^2\right)+\dfrac{a^2+b^2+c^2+2ab+2bc+2ac}{2}\right]=\)
\(=abc.\left[\dfrac{5}{2}.\left(a^2+b^2+c^2\right)+\dfrac{\left(a+b+c\right)^2}{2}\right]=\)
\(=abc.\dfrac{5}{2}.\left(a^2+b^2+c^2\right)\)
\(\Rightarrow\dfrac{a^5+b^5+c^5}{5}=abc.\dfrac{a^2+b^2+c^2}{2}\left(đpcm\right)\)
