# 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

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