Four real numbers cannot be distinct

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Four real numbers cannot be distinct

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Source: Spanish MO 2022 P2

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#1 h May 2, 2022, 6:14 PM

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Let $a,b,c,d$ be four positive real numbers. If they satisfy $a+b+\frac{1}{ab}=c+d+\frac{1}{cd}\quad\text{and}\quad\frac1a+\frac1b+ab=\frac1c+\frac1d+cd$ then prove that at least two of the values $a,b,c,d$ are equal.

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yobu

#2 h May 2, 2022, 6:55 PM

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Note that the polynomials $P(x)=(x-a)(x-b)(x-\frac{1}{ab})$ and $Q(x) = (x-c)(x-d)(x-\frac{1}{cd})$ are same because,
$P(x)=(x-a)(x-b)(x-\frac{1}{ab}) = x^3-(a+b+\frac{1}{ab})x^2+(\frac{1}{a}+\frac{1}{b}+ab)x-1 = x^3-(c+d+\frac{1}{cd})x^2+(\frac{1}{c}+\frac{1}{d}+cd)x-1 = Q(x)$ So we get the roots of both polynomials are same. In other words $\{a,b,\frac{1}{ab}\} = \{c,d,\frac{1}{cd}\}$ . Which implies that at least two of $a,b,c,d$ are equal. $\blacksquare$

This post has been edited 1 time. Last edited by yobu, May 2, 2022, 6:56 PM
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