Ta có a,b,c dương nên ta áp dụng Bđt Cô-si ta có:
\(a^3+b^3+c^3\ge3\sqrt[3]{a^3b^3c^3}=3abc\)
Dấu = khi \(a=b=c\)
Đpcm
\(a^3+b^3+c^3=3abc\)
\(\Rightarrow a^3+b^3+c^3-3abc=0\)
\(\Rightarrow\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc=0\)
\(\Rightarrow\left[\left(a+b\right)^3+c^3\right]-3ab\left(a+b+c\right)=0\)
\(\Rightarrow\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)=0\)
\(\Rightarrow\left(a+b+c\right)\left[\left(a+b\right)^2-c\left(a+b\right)+c^2-3ab\right]=0\)
\(\Rightarrow\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)=0\)
\(\Rightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)=0\)
\(\Rightarrow\left(a+b+c\right)\frac{1}{2}\left(2a^2+2b^2+2c^2-2ab-2ac-2bc\right)=0\)
\(\Rightarrow\left(a+b+c\right)\frac{1}{2}\left(a^2-2ab+b^2\right)\left(b^2-2bc+c^2\right)\left(c^2-2ac+a^2\right)=0\)
\(\Rightarrow\left(a+b+c\right)\frac{1}{2}\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0.\)
vì \(\left(a-b\right)^2\ge0\)
\(\left(b-c\right)^2\ge0\)
\(\left(c-a\right)^2\ge0\)
\(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
\(\Rightarrow a-b=b-c=c-a\)
\(\Rightarrow a=b=c\left(dpcm\right)\)
