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 $0 \le a,$ $b,$ $c,$ $d \le 1.$

Question

Let $0 \le a,$ $b,$ $c,$ $d \le 1.$ Find the possible values of the expression $$\sqrt{a^2 + (1 - b)^2} + \sqrt{b^2 + (1 - c)^2} + \sqrt{c^2 + (1 - d)^2} + \sqrt{d^2 + (1 - a)^2}.$$

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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!!

Answer 1

The max must be $4$ (while $a=b=c=d=1$) and min be $a=c$ and $b=d$).

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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.

Answer 2

Denote the expression by $P(a, b, c, d)$.

First, using Minkowski inequality, we have \begin{align*} P &\ge \sqrt{(a + b + c + d)^2 + (1 - b + 1 - c + 1 - d + 1 - a)^2}\\ &= \sqrt{x^2 + (4 - x)^2}\\ &= \sqrt{2(x - 2)^2 + 8}\\ &\ge 2\sqrt 2 \end{align*} where $x = a + b + c + d$. Also, $P(1/2, 1/2, 1/2, 1/2) = 2\sqrt 2$. Thus, the minimum of $P$ is $2\sqrt 2$.

Second, using $x + y + z + t \le \sqrt{4(x^2 + y^2 + z^2 + t^2)}$ for all $x, y, z, t \ge 0$ (well-known, the so-called AM-QM), we have \begin{align*} P &\le \sqrt{4[a^2 + (1 - b)^2 + b^2 + (1 - c)^2 + c^2 + (1 - d)^2 + d^2 + (1 - a)^2]}\\ &= \sqrt{16 - 8a - 8b - 8c - 8d + 8a^2 + 8b^2 + 8c^2 + 8d^2}\\ &\le 4 \end{align*} where we have used $a \ge a^2$ etc. Also, $P(1, 1, 1, 1) = 4$. Thus, the maximum of $P$ is $4$.

Thus, the range of $P$ is $[2\sqrt 2, 4]$.