Find $\max{\left(\min{\left(|a-b|,|b-c|,|c-d|,|d-e|,|e-a|\right)}\right)}$ on unit sphere

Question

Let $a,b,c,d,e\in \mathbb{R}$ such that $$a^2+b^2+c^2+d^2+e^2=1$$ find this value $$A=\max{\left(\min{\left(|a-b|,|b-c|,|c-d|,|d-e|,|e-a|\right)}\right)}$$ or more precisely, find $$\max_{a^2+b^2+c^2+d^2+e^2=1} \min(|a-b|, |b-c|, |c-d|, |d-e|, |e-a|)$$

I use computer have this $$A=\dfrac{2}{\sqrt{10}}$$

then equal holds if we suppose that $a\leq b\leq c\leq d\leq e$, then$$a=\frac{2}{\sqrt{10}},b=-\frac{1}{\sqrt{10}},c=\dfrac{1}{\sqrt{10}},d=-\frac{2}{\sqrt{10}},e=0$$

But I consider sometimes, I want use Cauchy-Schwarz inequality to solve it, I guess have this follow?

$$(|a-b|+|b-c|+|c-d|+|d-e|+|e-a|)^2\le(\dfrac{1}{10})(a^2+b^2+c^2+d^2+e^2)$$ then I can't. Thank you very much!

Answer 1

The situation described by the OP (situation (c) in the following figure) is optimal. This can be seen as follows:

We consider the dual problem instead: Five telescopic rods of length $\geq1$ each are hinged together at the ends so that a flexible and extensible pentagon is formed. This pentagon is then extended and squeezed to a "linear structure" $S$, i.e., such that the five hinges are in line. This structure is then positioned along the $x$-axis. Let $a_i$ $\>(1\leq i\leq 5)$ be the resulting positions of the five hinges. Our aim is to minimize the quantity $$\Phi:=\sum_{i=1}^5a_i^2\ .$$ A first step in this direction is to translate $S$ along the $x$-axis such that the centroid of the hinges is at the origin. The resulting $\Phi$-value will be denoted by $\Phi'(S)$ and can be considered as total moment of inertia of the five hinges with respect to their centroid $c=0$.

The "linear structure" $S$ has $2$ or $4$ return-hinges and the rest $180^\circ$-hinges. When there are $2$ return-hinges they can have $1$ or $2$ rods between them. These are the configurations (a) and (b) in the following figure. When there are $4$ return-hinges there is just one $180^\circ$-hinge, see (c) in the figure.

The red rods are critical: we have to ensure that their length is $\geq1$, whereas the black rods automatically have a length $>1$. The configurations (a)–(c) with all red rods of length 1 are possible and admissible. On the other hand, our physical intuition tells us that making any of these red rods longer than $1$ will increase the moment of inertia $\Phi'(S)$ of the configuration. In cases (b) and (c) some "internal shifting" is possible, but we know from experience that the symmetric situation has smallest moment of inertia. (I shall treat case (c) explicitly at the end.) It follows that the optimal positions in the three cases are given by $${\rm (a)}\qquad a_1=-2,\quad a_2=-1, \quad a_3=0, \quad a_4=1,\quad a_5=2,\qquad \Phi'=10\ ;$$ $${\rm (b)}\qquad a_1=-{3\over2},\quad a_2=-{1\over2}, \quad a_3={1\over2}, \quad a_4={3\over2},\quad a_5=0,\qquad \Phi'=5\ ;$$ $${\rm (c)}\qquad a_1=-1,\quad a_2=0, \quad a_3=1, \quad a_4=-{1\over2},\quad a_5={1\over2},\qquad \Phi'={5\over2}\ .$$ From this we conclude that the global minimum of $\Phi$, resp. $\Phi'$, is assumed in situation (c).

Returning to the original problem we therefore can say the following: When $\Phi=1$ is prescribed then at least one of the rods must have a length $$d\leq\sqrt{2\over 5}\ .$$ Update: There was some handwaving above. In the following I shall treat case (c) in detail, the other cases are simpler.

When the lengths $d_1$, $d_2$, $d_3$ of the three red rods are given then for suitable $u$, $v$ one has $$a_1=v-{2d_1+d_2\over3}, \quad a_2=v+{d_1-d_2\over 3},\quad a_3=v+{d_1+2d_2\over3},$$ $$ a_4=u-{d_3\over2},\quad a_5=u+{d_3\over2}\ .$$ From this one computes $$\Phi=3v^2+{2\over3}(d_1^2+d_1d_2+d_2^2)+2u^2+{1\over2}d_3^2\ .$$ It follows that for given $d_i$ $\>(1\leq i\leq 3)$ one may attain $$\Phi_{\min}={2\over3}(d_1^2+d_1d_2+d_2^2)+{1\over2}d_3^2\ ,$$ which is an increasing function of the $d_i$, as claimed in the main text.

Answer 2

The gist of the answer is that with every small but certain step the problem is simplified a little bit, hoping that in the end all of these steps together will lead to full comprehension.
Disclaimer: I think the proof is not yet completed by taking together all steps.

Step 1.The whole problem is invariant for cyclic permutations. Hence indeed, you cannot suppose without loss of generality that $a \le b \le c \le d \le e$, but you can suppose without loss of generality that $a$ is the smallest number and then investigate the remaining permutations, giving the following list of possibilities.

                    |a-b|,|b-c|,|c-d|,|d-e|,|e-a|
a ≤ b ≤ c ≤ d ≤ e     1     1     1     1     4
a ≤ b ≤ c ≤ e ≤ d     1     1     2     1     3
a ≤ b ≤ d ≤ c ≤ e     1     2     1     2     4
a ≤ b ≤ d ≤ e ≤ c     1     3     2     1     3
a ≤ b ≤ e ≤ c ≤ d     1     2     1     2     2
a ≤ b ≤ e ≤ d ≤ c     1     3     1     1     2
a ≤ c ≤ b ≤ d ≤ e     2     1     2     1     4
a ≤ c ≤ b ≤ e ≤ d     2     1     3     1     3
a ≤ c ≤ d ≤ b ≤ e     3     2     1     2     4
a ≤ c ≤ d ≤ e ≤ b     4     3     1     1     3
a ≤ c ≤ e ≤ b ≤ d     3     2     3     2     2     !(1)
a ≤ c ≤ e ≤ d ≤ b     4     3     2     1     2
a ≤ d ≤ b ≤ c ≤ e     2     1     2     3     4
a ≤ d ≤ b ≤ e ≤ c     2     2     3     2     3     !(2)
a ≤ d ≤ c ≤ b ≤ e     3     1     1     3     4
a ≤ d ≤ c ≤ e ≤ b     4     2     1     2     3
a ≤ d ≤ e ≤ b ≤ c     3     1     3     1     2
a ≤ d ≤ e ≤ c ≤ b     4     1     2     1     2
a ≤ e ≤ b ≤ c ≤ d     2     1     1     3     1
a ≤ e ≤ b ≤ d ≤ c     2     2     1     2     1
a ≤ e ≤ c ≤ b ≤ d     3     1     2     3     1
a ≤ e ≤ c ≤ d ≤ b     4     2     1     2     1
a ≤ e ≤ d ≤ b ≤ c     3     1     2     1     1
a ≤ e ≤ d ≤ c ≤ b     4     1     1     1     1

The numbers on the right are not so much the sizes of the intervals but they indicate how many intervals ( bounded by $a,b,c,d,e$ ) are in between ( if you understand what I mean ) . The two orderings marked with an exclamation sign (!) are the most promising, because the intervals to be minimized each contain at least two other intervals, which, so to speak, makes it more difficult to minimize them. The latter is desirable because what we seek in the end is a maximum.
Also note that the problem is invariant for reversal of the ordering. For example $a \le b \le c \le d \le e$ is the same as $a \ge b \ge c \ge d \ge e$ . Thus we see e.g. that the ordering assumed in the question is the one indicated as (1). There is more. If the order of (2) is reversed, then we get: $c \le e \le b \le d \le a$ . A cyclic permutation then gives: $a \le c \le e \le b \le d$ . Which is exactly case (1) again. We conclude that (1) is the only case we have to investigate, in the end. Thus, with no loss of generality: $$ a \le c \le e \le b \le d $$ Now take a look at the minmax function again. The arguments that have no chance to be minimal can be removed. Also note the disappearance of the absolute values: $$ A=\max{\left(\min{\left(|a-b|,|b-c|,|c-d|,|d-e|,|e-a|\right)}\right)} \qquad \Longleftrightarrow \qquad A=\max{\left(\min{\left(b-c,d-e,e-a\right)}\right)} $$

Step 2. Further information about the magnitude of the numbers can be obtained as follows. Let the numbers be diminished with an amount $\mu$ : $$ W(\mu) = (a-\mu)^2 + (b-\mu)^2 + (c-\mu)^2 + (d-\mu)^2 + (e-\mu)^2 $$ It is desirable that $W$ is minimal as a function of $\mu$ because that gives the least possible restriction on $(a,b,c,d,e)$ when $a^2+b^2+c^2+d^2+e^2 = 1$ . Differentiating gives, not much to our surprise, that the minimum of $W(\mu)$ is obtained for: $$ \frac{1}{2} \frac{d W}{d \mu} = a + b + c + d + e - 5 \mu = 0 \qquad \Longrightarrow \qquad \mu = \frac{a + b + c + d + e}{5} \qquad \Longrightarrow \qquad W(\mu) = a^2+b^2+c^2+d^2+e^2 - 5 \mu^2 $$ Thus, with no loss of generality, we can - and we must - choose our numbers $\; (a,b,c,d,e)$ , such that $a + b + c + d + e = 0$ . All choices in the question and the given answers and comments fulfill this condition, whether that is by coincidence or not ( I think not ).

Step 3. According to the above, the minmax function can now be split into three pieces: $$ (d-e \ge e-a \ge b-c) \vee (e-a \ge d-e \ge b-c) \quad \Longrightarrow \quad A = \max{(b-c)} \\ (d-e \ge b-c \ge e-a) \vee (b-c \ge d-e \ge e-a) \quad \Longrightarrow \quad A = \max{(e-a)} \\ (e-a \ge b-c \ge d-e) \vee (b-c \ge e-a \ge d-e) \quad \Longrightarrow \quad A = \max{(d-e)} $$ It is suggested herewith that we should anyway investigate the special case, where none of the intervals is the smallest: $$ e-a = d-e = b-c $$ Because only differences of the numbers $\; (a,b,c,d,e) \;$ are considered in the minmax function, there is still a degree of freedom in our problem, which is the choice of the origin. Therefore we shall assume in the sequel that $\; e = 0$ .
As far as the special case is considered, we are almost there: $$ -a = d = b-c \quad \Longrightarrow \quad a+b+c+d+e = b+c = 0 \quad \Longrightarrow \\ c = -b \quad ; \quad d = 2 b \quad ; \quad a = - 2 b $$ Giving: $$ a^2+b^2+c^2+d^2+e^2 = 2(2b)^2 + 2b^2 = 1 \quad \Longrightarrow \quad b = \frac{1}{\sqrt{10}} \quad \Longrightarrow \quad (a,b,c,d,e) = \frac{1}{\sqrt{10}} (-2,+1,-1,+2,0) \quad \Longrightarrow \quad A = \frac{2}{\sqrt{10}} $$ Which is exactly the solution as given in the question.

But why would the special case give the solution?

Step 4. Let's try to make a small modification of the special case, as follows. $$ a = -d = \lambda (b-c) \quad \Longrightarrow \quad c = -b \quad ; \quad d = 2 b \lambda \quad ; \quad a = - 2 b \lambda $$ Giving: $$ a^2+b^2+c^2+d^2+e^2 = 2(2b \lambda)^2 + 2 b^2 = 1 \quad \Longrightarrow \quad b = \frac{1}{\sqrt{8\lambda^2+2}} $$ If we suppose that $\; 0 < \lambda < 1$ : $$ A(\lambda) = \lambda (b-c) = \frac{2\lambda}{\sqrt{8\lambda^2+2}} $$ It follows with elementary algebra that $\; A(\lambda) < 2/\sqrt{10}$ .
Now suppose that $\; \lambda > 1$ , then: $$ A(\lambda) = (b-c) = \frac{2}{\sqrt{8\lambda^2+2}} $$ It follows with elementary algebra that, again, $\; A(\lambda) < 2/\sqrt{10}$ .

Step 5. Asymmetric case.
Let $a = -2 k$ , $c = -k$ , $e = 0$ , $b = \lambda k$ . Then: $$a+b+c+d+e=0 \quad \Longrightarrow \quad d = (3-\lambda) k$$ $$a^2+b^2+c^2+d^2+e^2=1 \quad \Longrightarrow \quad k=\frac{1}{\sqrt{2\lambda^2-6\lambda+14}}$$ Two cases are distinguished: $$ \lambda < 1 \quad \Longrightarrow \quad A(\lambda) = \max{(b-c)} = \frac{1+\lambda}{\sqrt{2\lambda^2-6\lambda+14}}$$ $$ \lambda > 1 \quad \Longrightarrow \quad A(\lambda) = \max{(d-e)} = \frac{3-\lambda}{\sqrt{2\lambda^2-6\lambda+14}}$$ We must prove that in both cases $A(\lambda) < 2/\sqrt{10}$ .
First case $\lambda < 1$ : $$ \frac{1+\lambda}{\sqrt{2\lambda^2-6\lambda+14}} < \frac{2}{\sqrt{10}} \quad \Longleftrightarrow \quad 10 (\lambda+1)^2 - 4 (2\lambda^2-6\lambda+14) < 0 \quad \Longleftrightarrow \quad 2 \lambda^2 + 44 \lambda - 46 < 0 \quad \Longleftrightarrow \quad (\lambda+11)^2-144 < 0 \quad \Longleftrightarrow \quad -23 < \lambda < 1 $$ This is well within the range $\; 0 < \lambda < 1 \;$ hence $\; A(\lambda) < 2/\sqrt{10}$ .
Second case $\lambda > 1$ : $$ \frac{3-\lambda}{\sqrt{2\lambda^2-6\lambda+14}} < \frac{2}{\sqrt{10}} \quad \Longleftrightarrow \quad 10 (\lambda-3)^2 - 4 (2\lambda^2-6\lambda+14) < 0 \quad \Longleftrightarrow \quad 2 \lambda^2 - 36 \lambda + 34 < 0 \quad \Longleftrightarrow \quad (\lambda-9)^2-64 < 0 \quad \Longleftrightarrow \quad 1 < \lambda < 17 $$ This is well within the range $\; 1 < \lambda < 3 \;$ hence $\; A(\lambda) < 2/\sqrt{10}$ .

Step 6. I should have read the question more carefully in the first place: a nice suggestion is there.
Because $\; A = \max{(\min{(e-a,d-e,b-c)})} \;$ and $\;a^2+b^2+c^2+d^2+e^2=1\;$ and $\;e=0\;$ it is clear that: $$ A \le \sqrt{\frac{(e-a)^2+(d-e)^2+(b-c)^2}{3}} = \sqrt{\frac{1-2\,b\,c}{3}} $$ Where the equality is valid only for the special case that $\; e-a=d-e=b-c $ . Thus in all other cases the inequality is strict: $$ \neg \; (\; e-a=d-e=b-c \; ) \quad \Longrightarrow \quad A < \sqrt{\frac{(e-a)^2+(d-e)^2+(b-c)^2}{3}} $$ Depending only on $\;b\;$ and $\;c\;$. This proves that our special case is optimal and that $\; A = 2/\sqrt{10} \;$, provided that a rather strange condition is fulfilled in addition to the strict inequality: $$ b\,c = - \frac{1}{10} $$

Step 7. Giving up. Because all of my steps toward full comprehension are clearly overruled by the answer with Christian Blatter as the author. In case you didn't notice, it's peanuts now to complete the proof. So give Christian a firm upvote, as I did!

Answer 3

You cannot suppose (without loss of generality) that $a\le b\le c\le d\le e$, because the quantity you're maximizing is not fully symmetric in the five variables. For example, taking $a=c=-\frac12$, $b=d=\frac12$, and $e=0$ shows that $A\ge\frac12$.

Answer 4

This is a comment that’s too long to fit in the usual format. I prove below the non-optimal bound $A\leq \sqrt{\frac{5+\sqrt{5}}{10}} \approx 0.8$. Let us denote

$$ x_1=x_6=a, x_2=b, x_3=c, x_4=d, x_5=e $$

Then, we have

$$ \big(\frac{5+\sqrt{5}}{2}\big) \bigg(\sum_{k=1}^5 x_k^2\bigg)- \bigg(\sum_{k=1}^5 (x_k-x_{k+1})^2\bigg)= \big(\frac{\sqrt{5}+1}{2}\big) \bigg(x_5+\frac{(\sqrt{5}-1)(x_1+x_4)}{2}\bigg)^2+ \bigg(x_3+x_4-\frac{(\sqrt{5}-1)x_1}{2}\bigg)^2+ \big(\frac{\sqrt{5}-1}{2}\big) \bigg(x_1+x_3+\frac{(\sqrt{5}+1)x_2}{2}\bigg)^2 $$

We deduce that

$$ \big(\frac{5+\sqrt{5}}{2}\big) \bigg(\sum_{k=1}^5 x_k^2\bigg) \geq \bigg(\sum_{k=1}^5 (x_k-x_{k+1})^2\bigg) $$

So if we put $\varepsilon={\sf min}(|x_{k+1}-x_k|)$, we have

$$ \frac{5+\sqrt{5}}{2} \geq \sum_{k=1}^5 (x_k-x_{k+1})^2 \geq 5\varepsilon^2 $$

and hence

$$ \varepsilon \leq \sqrt{\frac{5+\sqrt{5}}{10}} $$

Answer 5

Remark: Here is a solution without calculus.

When $a = \frac{\sqrt{10}}{10}, b = - \frac{\sqrt{10}}{10}, c = \frac{\sqrt{10}}{5}, d = 0, e = - \frac{\sqrt{10}}{5}$, we have $$\min(|a-b|, |b-c|, |c-d|, |d-e|, |e-a|) = \frac{\sqrt{10}}{5}.$$

We claim that the maximum is $\frac{\sqrt{10}}{5}$.

It suffices to prove that, for all reals $a^2 + b^2 + c^2 + d^2 + e^2 = 1$, $$f(a, b, c, d, e) := \min(|a-b|, |b-c|, |c-d|, |d-e|, |e-a|) \le \frac{\sqrt{10}}{5}.$$

Clearly, we only need to prove the case that $(a-b)(b-c)(c-d)(d-e)(e-a) \ne 0$.

Note that the inequality does not change if $(a, b, c, d, e)$ is substituted with $(-a, -b, -c, -d, -e)$. Thus, we may assume that at most two of $a-b, b-c, c-d, d-e, e-a$ are negative. Since the inequality is cyclic, we may assume that $e - a < 0$.

We split into five cases.

Case 1: $a - b, b - c, c - d, d- e \ge 0$

If $f > \sqrt{10}/5$, we have $a - e = (a-b) + (b-c) + (c-d) + (d-e) > 4\sqrt{10}/5$ which contradicts $a - e \le \sqrt{2(a^2 + e^2)} \le \sqrt 2$.

$\phantom{2}$

Case 2: $a - b < 0$ and $b - c, c - d, d - e \ge 0$

If $f > \sqrt{10}/5$, we have $b - e = (b-c) + (c-d) + (d - e) > 3\sqrt{10}/5$ which contradicts $b - e \le \sqrt{2(b^2 + e^2)} \le \sqrt 2$.

$\phantom{2}$

Case 3: $d - e < 0$ and $a - b, b - c, c - d \ge 0$

If $f > \sqrt{10}/5$, we have $a - d = (a - b) + (b - c) + (c - d) > 3\sqrt{10}/5$ which contradicts $a - d \le \sqrt{2(a^2 + d^2)} \le \sqrt 2$.

$\phantom{2}$

Case 4: $b - c < 0$ and $a-b, c- d, d - e \ge 0$

We have \begin{align*} f &= \min(|a-b|, |b-c|, |c-d|, |d-e|, |e-a|)\\ &\le \min(|a-b|, |c-d|, |d-e|)\\ &= \min(a-b, c - d, d - e)\\[6pt] &\le \frac15(a - b) + \frac25(c-d) + \frac25(d - e)\\[6pt] &= \frac15(a - b + 2c - 2e)\\[6pt] &\le \frac15\sqrt{(a^2 + b^2 + c^2 + e^2)(1 + 1 + 2^2 + 2^2)} \\[6pt] &\le \frac{\sqrt{10}}{5}. \end{align*}

$\phantom{2}$

Case 5: $c - d < 0$ and $a-b, b-c, d-e\ge 0$

We have \begin{align*} f &= \min(|a-b|, |b-c|, |c-d|, |d-e|, |e-a|)\\ &\le \min(|a-b|, |b-c|, |d-e|)\\ &= \min(a-b, b - c, d - e)\\[6pt] &\le \frac25(a - b) + \frac25(b-c) + \frac15(d - e)\\[6pt] &= \frac15(2a - 2c + d - e)\\[6pt] &\le \frac15\sqrt{(a^2 + c^2 + d^2 + e^2)(2^2 + 2^2 + 1 + 1)} \\[6pt] &\le \frac{\sqrt{10}}{5}. \end{align*}

We are done.