# An Inequality from Marocco, with a Proof, or Is It?

### Statement

$1\le a,b,c,d\le 2.$ Prove that

$4|(a-b)(b-c)(c-d)(d-a)|\le abcd.$

### Solution

Without lost of generality, we may assume $a\le b\le c\le d.$ If a pair of (cyclically) successive numbers are equal, the inequality obviously holds. So assume $1\le a\lt b\lt c\lt d\le 2.$

Consider the polynomial $f(x)=(a-x)(x-c).$ It attains its maximum at $\displaystyle x=\frac{a+c}{2}$ which is equal to $\displaystyle f(\frac{a+c}{2})=\frac{(c-a)^2}{4}.$ This is the maximal value $|f(x)|$ attains in the interval $[a,c].$ Since $b\in (a,c),$ it follows that $\displaystyle |(a-b)(b-c)|\le\frac{(c-a)^2}{4}\le\frac{1}{4}.$

On the other hand, the difference of any two numbers in the interval $[1,2]$ does not exceed $1,$ which implies $|(c-d)(d-a)|\le 1.$ Combinting the two inequalities we obtain

$\displaystyle |(a-b)(b-c)(c-d)(d-a)|\le\frac{1}{4}\cdot 1\le\frac{abcd}{4}.$

### What's wrong?

The problem imposes a certain order on the numbers $a,b,c,d$ that assumes nothing about their magnitudes. The additional assumption $a\le b\le c\le d$ may be simply not true (which in fact what happens).

### Acknowledgment

Leo Giugiuc has kindly posted the problem and two solutions. One of his and Dan Sitaru (Solution 1) and that by Amine Idirissi from Morocco (Solution 2), with the remark that the problem was offered at the the Morroco Mathematical Olympiad. The two solutions can be found on a separate page.