<s>[INST] 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 $x=-\frac{1}{2}$
 [/INST] 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 $$\frac{x}{1-x}=\frac{1-x}{1+x}\Rightarrow x^2-2x+1=x^2+x\Rightarrow x=\frac{1}{3}$$
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