# 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.

This is best explained with a picture:

The length you have on the LHS is the length of the dotted red path in the unit square.

Now use some reflections in order to "straighten" that path. You will get:

$$\sqrt{a^2+(1-b)^2}+\sqrt{b^2+(1-c)^2}+\sqrt{c^2+(1-a)^2}\geq \sqrt{(1+a)^2+(2-a)^2}$$

and now it is trivial that the RHS has a minimum for $a=\frac{1}{2}$, whose value is $\frac{3}{2}\sqrt{2}$.

Another option is to use the QM-AM (also known as Cauchy-Schwarz) inequality as orangeskid did. With the previous line, we can also improve the original inequality up to:

$$\sqrt{a^2+(1-b)^2}+\sqrt{b^2+(1-c)^2}+\sqrt{c^2+(1-a)^2}\geq \sqrt{2}\left(\frac{3}{2}+\frac{\left|a-\frac{1}{2}\right|^2}{3+\left|a-\frac{1}{2}\right|}\right).$$

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}

Hint:

$$\sqrt{\frac{a^2+(1-b)^2}{2}}\ge \frac{|a| + |1-b|}{2}$$

Add up, RHS $\ge \frac{|a|+|1-a|+|b|+|1-b|+|c|+|1-c|}{2}\ge \frac{3}{2}$, on LHS it's our sum divided by $\sqrt{2}$.

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!