8

let $a,b,c,d,e\in [0,1]$, show that $$\sqrt{|a-b|}+\sqrt{|b-c|}+\sqrt{|c-d|}+\sqrt{|d-e|}+\sqrt{|e-a|}\le 3+\sqrt{2}$$

I have tried to use AM-GM inequality, but get no result as follows: $$(|a-b|+|b-c|+|c-d|+|d-e|+|e-a|)(1+1+1+1)\ge (\sqrt{|a-b|}+\sqrt{|b-c|}+\sqrt{|c-d|}+\sqrt{|d-e|}+\sqrt{|e-a|})^2$$

1 Answers1

5

Five is an odd number, therefore the differences $$ b - a, c - b, d - c, e - d, a - e $$ (when viewed as a cyclic sequence) cannot have only alternating signs. There must be two successive differences which are both $\le 0$ or both $\ge 0$. Therefore in $$ a, b, c, d, e $$ (also viewed as a cyclic sequence) there must be three successive numbers which are increasing or decreasing.

Without loss of generality, we can assume that $a \le b \le c$. It is easy to calculate that $$ f(x) = \sqrt{x-a} + \sqrt{c - x} \, , \, a \le x \le c $$ attains its maximum at the midpoint $(a+c)/2$ of the interval and therefore $$ \sqrt{|a-b|}+\sqrt{|b-c|} = \sqrt{b-a} + \sqrt{c - b} \le 2 \sqrt{\frac{b-a}{2}} \le \sqrt 2 \, . $$ Since $$ \sqrt{|c-d|}+\sqrt{|d-e|}+\sqrt{|e-a|} \le 1 + 1 + 1 = 3$$ holds trivially, the conclusion follows.

Equality holds for $$ (a, b, c, d, e) = (0, \frac 12, 1, 0, 1) \, $$ so the bound $3 + \sqrt 2$ is sharp.

Martin R
  • 113,040
  • One question, could this be treated as a five variable function and use Lagrange multipliers to find its maximum in the region considered? – YoTengoUnLCD Sep 08 '15 at 21:04
  • @YoTengoUnLCD: It might be at bit cumbersome, because the absolute value function is not differentiable everywhere. Also the example shows that the maximum is attained at the boundary of the domain. – But it may be possible of course, I was just happy that I found a solution at all :) – Martin R Sep 08 '15 at 21:14