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.

By Minkowski's inequality,
\begin{align}&\sqrt{a^2 + (1 - b)^2} + \sqrt{b^2 + (1 - c)^2} + \sqrt{c^2 + (1 - a)^2} \\
&= \|(a,1-b)\|_2 + \|(b,1-c)\|_2 + \|(c,1-a)\|_2\\
&= \frac{\|(a,1-b)\|_2 + \|(1 - c,b)\|_2}{2} + \frac{\|(b,1-c)\|_2 + \|(1-a,c)\|_2}{2}\\
& + \frac{\|(c,1-a)\|_2 + \|(1-b, a)\|_2}{2}\\
&\ge \frac{\|(a + 1 - c, 1)\|_2}{2} + \frac{\|(b + 1 - a, 1)\|_2}{2} + \frac{\|(c + 1 - a, 1)\|_2}{2}\\
&\ge \frac{\|(a + 1 - c) + (b + 1 - a) + (c + 1 - a),3)\|_2}{2}\\
&= \frac{\|(3,3)\|_2}{2}\\
&= \frac{3\sqrt{2}}{2}.
\end{align}
