Show that three numbers form an arithmetic progression 
The numbers $a,b$ and $c$ form  an arithmetic progression. Show that the numbers $a^2+ab+b^2,a^2+ac+c^2,b^2+bc+c^2$ also form an arithmetic progression.

We have that $2b=a+c$ (we know that a sequence is an arithmetic progression iff $a_n=\dfrac{a_{n-1}+a_{n+1}}{2}\text{   } \forall \text{   }n\ge2$). I am stuck here and I would be very grateful if you could give me a hint.

More hint:
\begin{align}&\quad(a^2+ab+b^2) + (b^2+bc + c^2) \\&= a^2+c^2 +2b^2 + b(a+c) \\&= a^2 + c^2 + 2b^2 + b(2b) \\&= a^2 + c^2 + 4b^2\\&=a^2+c^2+(a+c)^2\end{align}
In the worst case you can just substitute $b = \frac {a+c}2$.
