Minimum of $\sqrt{a^2+b^2}+\sqrt{(a-1)^2+b^2}+\sqrt{a^2+(b-1)^2}+\sqrt{(a-1)^2+(b-1)^2}.$

Question

Let $a$ and $b$ be real numbers. Find the minimum of $$\sqrt{a^2+b^2}+\sqrt{(a-1)^2+b^2}+\sqrt{a^2+(b-1)^2}+\sqrt{(a-1)^2+(b-1)^2}.$$

The problem is from an inequality book. While I was doing the problem myself, I got the following results:

From Minkowski's inequality, we have
$\sqrt{a^2+b^2}+\sqrt{(a-1)^2+b^2}+\sqrt{a^2+(b-1)^2}+\sqrt{(a-1)^2+(b-1)^2}\\ \geq \sqrt{(a+a-1+a+a-1)^2+(b+b-1+b+b-1)^2}\\ =\sqrt{(4a-2)^2+(4b-2)^2}\\ \geq \sqrt{8(2a-1)(2b-1)} \ \ \ \ \ \ \text{[From AM-GM]}$

Equality holds when $4a-2=4b-2\implies a=b$. And the minimum $\sqrt{8(2a-1)(2b-1)}=0$ when $a=b=\frac 12$.
Here, everything seemed right to me. But when I plugged in $a=b=\frac 12$ in the original expression, I didn't get the minimum $0$ rather I got $2\sqrt 2$.

When I checked the solution in the book, a geometric solution was given and the minimum was indeed $2\sqrt 2$. So, I couldn't find where my mistake is. It will be helpful for me if someone can find that. Other algebraic solutions are also welcome.
I'm not good at inequalities, so pardon silly mistakes. Thanks.

Answer 1

It seems to me that you misapply Minkowski's inequality. More accurately: \begin{align*} &\sqrt{a^2+b^2}+\sqrt{(a-1)^2+b^2}+\sqrt{a^2+(b-1)^2}+\sqrt{(a-1)^2+(b-1)^2}\\ &\geq\sqrt{2a^2+2(a-1)^2+2b^2+2(b-1)^2}\\ &=2\sqrt{\left(a-\frac{1}{2}\right)^2+\left(b-\frac{1}{2}\right)^2+\frac{1}{2}}.\end{align*} Then it's just like yours.

Answer 2

Assuming you got the minimum $0$ and you wanna ask what is your mistake.

Then each term reduces to $\sqrt{\dfrac{1}{2}}$

So $\sqrt{\dfrac{1}{2}}+\sqrt{\dfrac{1}{2}}+\sqrt{\dfrac{1}{2}}+\sqrt{\dfrac{1}{2}} = \sqrt{2}$

Just to help you help here is another helpful solution-

In the cartesian system of coordinates consider the points: $O(0,0)$, the origin; $A(x,y);B(x,y-1);C(x-1,y-1);D(x-1,y)$. $A,B,C,D$ are the vertices of a square with the side $1$. Results: $AC=BD=\sqrt2$. Denote $M\left(x-\dfrac{1}{2},y-\dfrac{1}{2}\right)=AC\cap BD$ the center of the square.

$\sqrt{x^2+y^2}=OA; \sqrt{x^2+(y-1)^2}=OB; \sqrt{(x-1)^2+(y-1)^2}=OC; \sqrt{(x-1)^2+y^2}=OD$. Using the triangle inequality: $\sqrt{x^2+y^2}+\sqrt{(x-1)^2+(y-1)^2}=OA+OC\ge AC=\sqrt2$, with equality for $O\in[AC]$. $\sqrt{x^2+(y-1)^2}+\sqrt{(x-1)^2+y^2}=OB+OD\ge BD=\sqrt2$, with equality for $O\in[BD]$.

Results: $\sqrt{x^2+y^2}+\sqrt{(x-1)^2+(y-1)^2}+\sqrt{x^2+(y-1)^2}+\sqrt{(x-1)^2+y^2}\ge2\sqrt2$, with equality for $O\equiv M\Longleftrightarrow x=y=\dfrac{1}{2}$.