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 $x=-\frac{1}{2}$
 A: 
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 $E$.
A: Using Minkowski Inequality
$$\sqrt{a^2+b^2}+\sqrt{c^2+d^2}\geq \sqrt{(a+c)^2+(b+d)^2}$$
and equality hold when $\displaystyle \frac{a}{b} = \frac{c}{d}$
So $$\sqrt{x^2+(1-x)^2}+\sqrt{(1-x)^2+(1+x)^2}\geq \sqrt{[x+(1-x)]^2+[(1-x)+(1+x)]^2}=\sqrt{5}$$
and Equality hold when $$\frac{x}{1-x}=\frac{1-x}{1+x}\Rightarrow x^2-2x+1=x^2+x\Rightarrow x=\frac{1}{3}$$
A: A posible way: Let $A,B,C$ a triangle and $AD$ the height. The problem is $AD=1-x$, $BD=x$, $CD=1+x$ and you want minimize $BA+AC$, but note that the area is fixed ($(1-x)(2x+1)/2$), then this sum is minime if the triangle is right in A. 
A: Write $f(x)=g(x)+h(x)$. Now $g(x)$ has minima at $x=1/2$ and $h(x)$ is increasing function with minima at $x=0$.Then $f(x)$ has minima at ?
