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Part of 2018 Benelux

Problems(1)

Benelux Olympiad 2018, Problem 1

Source: Benelux 2018, Problem 1

4/28/2018
(a) Determine the minimal value of (x+1y)(x+1y−2018)+(y+1x)(y+1x−2018),\displaystyle\left(x+\dfrac{1}{y}\right)\left(x+\dfrac{1}{y}-2018\right)+\left(y+\dfrac{1}{x}\right)\left(y+\dfrac{1}{x}-2018\right), where xx and yy vary over the positive reals.
(b) Determine the minimal value of (x+1y)(x+1y+2018)+(y+1x)(y+1x+2018),\displaystyle\left(x+\dfrac{1}{y}\right)\left(x+\dfrac{1}{y}+2018\right)+\left(y+\dfrac{1}{x}\right)\left(y+\dfrac{1}{x}+2018\right), where xx and yy vary over the positive reals.
BxMOBeneluxinequalitiesalgebra