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.

Note that
$$a^2+b^2+c^2-\frac{1}{3}(a+b+c)^2=\frac{1}{3}((a-b)^2+(b-c)^2+(a-c)^2)$$
(this is useful if you read Mark Bennet's hint). 
So it remains to prove that
$$-2(a-b+1)\leq \frac{1}{3}((a-b)^2+(b-c)^2+(a-c)^2).$$
Now you may use the real variables $x=a-b$, $y=b-c$ and $z=c-a$ (and you are right, the condition $a>1$, $b>1$ and $c>1$ is irrelevant!):
$$-6x-6\leq x^2+y^2+z^2.$$
That is equivalent to
$$\sqrt{3}\leq \sqrt{(x+3)^2+y^2+z^2}$$
where $x+y+z=0$. The above inequality holds because  the distance of any point on the plane $x+y+z=0$ from the point $(-3,0,0)$ is greater or equal to $\sqrt{3}$.
