Key idea. Expand both sides and use the fact that real squares are non-negative.

1. The left side expands to

   \[2a^2+2b^2+2c^2+2ab+2bc+2ac.\]

2. The right side expands to

   \[a^2+b^2+c^2+2ab+2bc+2ac.\]

3. Cancel the same mixed terms on both sides, then subtract \(a^2+b^2+c^2\). The result is \(a^2+b^2+c^2=0\).

4. Each square is at least zero. If even one were positive, their sum would be positive. Therefore all three squares, and all three numbers, must be zero.

5. Substituting \(a=b=c=0\) makes both original sides zero, so this triple really works.

Remember. A sum of real squares is zero only when every square is zero.