<s>[INST] Range of $\sqrt{a^2 + (1 - b)^2} + \sqrt{b^2 + (1 - c)^2} + \sqrt{c^2 + (1 - d)^2} + \sqrt{d^2 + (1 - a)^2}$ on $0 \le a,$ $b,$ $c,$ $d \le 1.$ Let $0 \le a,$ $b,$ $c,$ $d \le 1.$ Find the possible values of the expression
$$\sqrt{a^2 + (1 - b)^2} + \sqrt{b^2 + (1 - c)^2} + \sqrt{c^2 + (1 - d)^2} + \sqrt{d^2 + (1 - a)^2}.$$

I tried to use some inequalities to find the bounds of the expression, but it didn't really work. Also, I don't know calculus yet, so please keep the responses and hints non-calc.
Thanks in advance!!
 [/INST] 
The max must be $4$ (while $a=b=c=d=1$) and min be $a=c$ and $b=d$).

Define
$$L(a_1,a_2,b_1,b_2,c_1,c_2,d_1,d_2,\lambda_1,\lambda_2,\lambda_3,\lambda_4)=\sqrt{a_1^2+b_2^2}+\sqrt{b_1^2+c_2^2}+\sqrt{c_1^2+d_2^2}+\sqrt{d_1^2+a_2^2}+\lambda_1(a_1+a_2-1)+\lambda_2(b_1+b_2-1)+\lambda_3(c_1+c_2-1)+\lambda_4(d_1+d_2-1).$$
Let $\nabla L=0$, we find $a_1=b_2=c_1=d_2,a_2=b_1=c_2=d_1$ are  stationary points of L. Hence it could be the extremum of the function in question.
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