An Inequality from Marocco


$1\le a,b,c,d\le 2.$ 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:

  1. $x,y,z\lt 1\lt t:$

    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

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

  2. $t\lt\ 1\lt x,y,z:$

    This case reduces to the previous by inverting the four fractions that we defined.

  3. $x,y\lt 1\lt z,t:$

    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

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


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 reamrk that the problem was offered at the the Morroco Mathematical Olympiad. There is also one invalid solution that conceals a misleading argument. You may want to locate it.


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