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.
 A: 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 $\sqrt{3}$.
A: Let $a=x+1$, $b=y+1$ and $c=z+1$.
Thus, $x$, $y$ and $z$ are positives and we need to prove that:
$$3((x+1)^2+(y+1)^2+(z+1)^2+2(x-y+1))-(x+y+z+3)^2\geq0$$ or
$$x^2+y^2+z^2-xy-xz-yz+3x-3y+3\geq0$$ or
$$z^2-(x+y)z+x^2+y^2-xy+3x-3y+3=0,$$ for which it's enough to prove that 
$$(x+y)^2-4(x^2+y^2-xy+3x-3y+3)\leq0$$ or
$$(x-y+2)^2\geq0.$$
We see that your inequality is true for all real variables.
A: 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 $$(A+B+C)^2=A^2+B^2+C^2+2AB+2BC+2AC\le 3(A^2+B^2+C^2)$$
