<s>[INST] Prove that $\sqrt{a^2+(1-b)^2}+\sqrt{b^2+(1-c)^2}+\sqrt{c^2+(1-a)^2}\geq\frac{3\sqrt{2}}{2}$ 
Prove that the following inequality
  $$\sqrt{a^2+(1-b)^2}+\sqrt{b^2+(1-c)^2}+\sqrt{c^2+(1-a)^2}\geq\frac{3\sqrt{2}}{2}$$
  holds for arbitrary real numbers $a$, $b$ and $c.$

Someone says, "It's very easy problem. It can also be proved by AM–GM inequality." But I can't reason out the answer to this question.
 [/INST] By C-S
$$\left(\sum_{cyc}\sqrt{a^2+(b-1)^2}\right)^2=\sum_{cyc}\left(a^2+(b-1)^2+2\sqrt{(a^2+(b-1)^2)(b^2+(c-1)^2}\right)\geq$$
$$\geq\sum_{cyc}(a^2+b^2-2b+1+2(ab+(b-1)(c-1)))=$$
$$=\sum_{cyc}(2a^2-2a+1+2ab+2ab-4a+2)=$$
$$=\sum_{cyc}(2a^2+4ab-6a+3)=2(a+b+c)^2-6(a+b+c)+9.$$
Thus, it remains to prove that
$$2(a+b+c)^2-6(a+b+c)+9\geq\frac{9}{2}$$ or
$$4(a+b+c)^2-12(a+b+c)+9\geq0$$ or
$$(2(a+b+c)-3)^2\geq0.$$
Done!
</s>