# An Inequality from Marocco

### Statement

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

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

### Solution 1

Set $\displaystyle x=\frac{a}{b},\;y==\frac{b}{c},\;z=\frac{c}{d},\;t=\frac{d}{a}.$ From the defintion $\displaystyle\frac{1}{2}\le x,y,z,t\le 2$ and $xyzt=1.$ Under this condition we need to prove that

$\displaystyle |(1-x)(1-y)(1-z)(1-t)|\le\frac{1}{4}.$

If $1\in\{x,y,z,\},$ we are done. Otherwise, we shall consider three cases:

**$x,y,z\lt 1\lt t:$**We have $\displaystyle 1-x\le\frac{1}{2},$ $\displaystyle 1-y\le\frac{1}{2},$ $\displaystyle 1-z\le\frac{1}{2},$ $|1-t|\le 1,$ implying

$\displaystyle |(1-x)(1-y)(1-z)(1-t)|\le\frac{1}{8}\lt\frac{1}{4}.$

**$t\lt\ 1\lt x,y,z:$**This case reduces to the previous by inverting the four fractions that we defined.

**$x,y\lt 1\lt z,t:$**We have $\displaystyle 1-x\le\frac{1}{2},$ $\displaystyle 1-y\le\frac{1}{2},$ $|1-z|\le 1,$ $|1-t|\le 1,$ implying

$\displaystyle |(1-x)(1-y)(1-z)(1-t)|\le\frac{1}{4}.$

The equality clearly holds when two of them are $\displaystyle\frac{1}{2}$ and the other two equal $2.$ In the original notations we have two solutions $(1,2,1,2)$ and $(2,1,2,1).$

### Solution 2

From $\displaystyle\frac{1}{2}\le x,y,z,t\le 2$ we get $\displaystyle (1-x)^{2}\le\frac{x}{2},$ $\displaystyle (1-y)^{2}\le\frac{y}{2},$ $\displaystyle (1-z)^{2}\le\frac{z}{2},$ $\displaystyle (1-t)^{2}\le\frac{t}{2},$ multiplying which we get

$\displaystyle |(1-x)(1-y)(1-z)(1-t)|\le\sqrt{\frac{xyzt}{16}}=\frac{1}{4}.$

### Acknowledgment

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