Giải:
Ta có:
\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=4\left(a^2+b^2+c^2-ab-bc-ac\right)\)
\(\Leftrightarrow a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ca+a^2=4a^2+4b^2+4c^2-4ab-4bc-4ac\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca=4a^2+4b^2+4c^2-4ab-4bc-4ac\)
\(\Leftrightarrow2\left(a^2+b^2+c^2-ab-bc-ca\right)=4\left(a^2+b^2+c^2-ab-bc-ac\right)\)
\(\Leftrightarrow\left(a^2+b^2+c^2-ab-bc-ca\right)=2\left(a^2+b^2+c^2-ab-bc-ac\right)\)
\(\Rightarrowđpcm\)
Ta có:
\(\left(a-b\right)^2\)+\(\left(b-c\right)^2\)+\(\left(c-a\right)^2=4\left(a^2+b^2+c^2-ab-bc-ca\right)\)
\(\Leftrightarrow a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ca+a^2=4\left(a^2+b^2+c^2-ab-bc-ca\right)\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca=4\left(a^2+b^2+c^2-ab-bc-ac\right)\)
\(\Leftrightarrow2\left(a^2+b^2+c^2-ab-bc-ac\right)=4\left(a^2+b^2+c^2-ab-bc-ca\right)\)
\(\Leftrightarrow-2\left(a^2+b^2+c^2-ab-bc-ac\right)=0\)
\(\Leftrightarrow a^2+b^2+c^2-ab-bc-ac=0\)
\(\Leftrightarrow a=b=c\left(đpcm\right)\)
