Minimum value of $f(x) = \sqrt{x^2 + (1-x)^2} + \sqrt{(1-x)^2 +(1+x)^2}$ If  $f(x) = \sqrt{x^2 + (1-x)^2} + \sqrt{(1-x)^2 +(1+x)^2}$
Find the minimum value of the function
I tried using the AMGM inequality and differentiation but didn't know how to solve it any ideas?
This is from a math competition. ( I would like to see the most efficient way as I think differentiation in a math competition is not that efficient)
Using Mogjals comment and using the AM-GM inequality , setting $A = \sqrt{x^2 + (1-x)^2}$ and $B=\sqrt{(1-x)^2 +(1+x)^2}$
then
$A+B \geq 2\sqrt{AB}$
$$  \sqrt{x^2 + (1-x)^2} + \sqrt{(1-x)^2 +(1+x)^2} \geq 2\sqrt{2}\sqrt{x^2+1}\sqrt{2x^2-2x+1}$$
With equality if and only $A=B$ so $x=-\frac{1}{2}$


Let $A=(0,1)$, $B=(1,1)$, and $C=(-x,x)$ as in the picture. Then 
$$AC=\sqrt{x^2+(1-x)^2}, BC=\sqrt{(1+x)^2+(1-x)^2}.$$
Let $D$ be symmetric to $A$ about $x+y=0$ (trajectory of $C$). Apparently
$$AC+BC\ge BD=AE+BE.$$
Minimal is attained at $C=E$. I leave you figure the coordinates of $E$.
