Prove $\frac{1}{3}(a+b+c)^2\leq a^2 + b^2 + c^2 + 2(a-b+1).$ 
Prove that for $a>1$,$b>1$ and $c>1$ where $a,b,c\in \mathbb{R}$
$$\frac{1}{3}(a+b+c)^2\leq a^2 + b^2 + c^2 + 2(a-b+1).$$

My attempt: it is not so clear why is $a>1$, $b>1$ and $c>1$, but I factor left side, and all put ond side to prove $\geq 0$, but it is not an easy inequality.

Well here is a start. Set $A=a+1$ and $B=b-1$ and $C=c$ and the equation becomes $$\frac 13(A+B+C)^2\le A^2+B^2+C^2$$
Now consider $$(A-B)^2+(B-C)^2+(C-A)^2\ge 0$$ so that $$2A^2+2B^2+2C^2-2AB-2BC-2AC\ge 0$$ so that $$AB+BC+AC\le A^2+B^2+C^2$$
Now $$(A+B+C)^2=A^2+B^2+C^2+2AB+2BC+2AC\le 3(A^2+B^2+C^2)$$
