\(\text{Ta có : }a^3+b^3+c^3-3abc=0\\ \Rightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)=0\\ Mà\text{ }a;b;c>0\left(a;b;c\text{ }là\text{ }3\text{ }cạnh\text{ }của\text{ }1\text{ }\Delta\right)\\ \Rightarrow a^2+b^2+c^2-ab-ac-bc=0\\ \Rightarrow2\left(a^2+b^2+c^2-ab-ac-bc\right)=0\\ \Rightarrow2a^2+2b^2+2c^2-2ab-2ac-2bc=0\\ \Rightarrow\left(a^2-2ab+b^2\right)+\left(a^2-2ac+c^2\right)+\left(b^2-2bc+c^2\right)=0\\ \Rightarrow\left(a-b\right)^2+\left(a-c\right)^2+\left(b-c\right)^2=0\\ Do\text{ }\left(a-b\right)^2\ge0\forall a;b\\ \left(a-c\right)^2\ge0\forall a;c\\ \left(b-c\right)^2\ge0\forall b;c\\ \Rightarrow\left(a-b\right)^2+\left(a-c\right)^2+\left(b-c\right)^2\ge0\forall a;b;c\\ Dấu\text{ }"="\text{ }xảy\text{ }ra\text{ }khi:\left\{{}\begin{matrix}\left(a-b\right)^2=0\\\left(a-c\right)^2=0\\\left(b-c\right)^2=0\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}a-b=0\\a-c=0\\b-c=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=b\\a=c\\b=c\end{matrix}\right.\Rightarrow a=b=c\\ \Rightarrow\Delta\text{ }có\text{ }các\text{ }cạnh\text{ }a;b;c\text{ }là\text{ }\Delta\text{ }đều\text{ }\left(Định\text{ }nghĩa\text{ }\Delta\text{ }đều\text{ }\right)\)
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