<s>[INST] 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.
 [/INST] 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.
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