<s>[INST] How find this inequality minimum $\sum\limits_{cyc}\sqrt{a^2+b^2+ab-2a-b+1}$ 
Let $0<a,b,c<1$, find this follow minimum
  $$\sqrt{(a+b)^2-(a+1)(b+2)+3}+\sqrt{(b+c)^2-(b+1)(c+2)+3}+\sqrt{(c+a)^2-(c+1)(a+2)+3}$$

My try:
since
$$(a+b)^2-(a+1)(b+2)+3=a^2+b^2+2ab-ab-2a-b-2+3=a^2+b^2+ab-2a-b+1$$
so we only find this follow minimum 
$$\sum_{cyc}\sqrt{a^2+b^2+ab-2a-b+1}$$
But I can't.Thank you 
 [/INST] $a^2+b^2+ab-2a-b+1=(b+\dfrac{a-1}{2})^2+\dfrac{3}{4}(a-1)^2$
$\sum\limits_{cyc}\sqrt{a^2+b^2+ab-2a-b+1} \ge \sqrt{(\sum\limits_{cyc} (b+\dfrac{a-1}{2})^2+\dfrac{3}{4}(\sum\limits_{cyc}(a-1))^2}=\sqrt{(\dfrac{3}{2}(\sum\limits_{cyc} a-1)^2+\dfrac{3}{4}(\sum\limits_{cyc} a-3)^2}=\sqrt{3((\sum\limits_{cyc} a)^2-3\sum\limits_{cyc} a+3)}\ge \dfrac{3}{2}$
when $\sum\limits_{cyc} a=\dfrac{3}{2}$ and $ a=b=c=\dfrac{1}{2}$ get min.
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