Question by Alone - Discuss with MathYouLike

Alone

27/12/2017 at 22:04

Let x,y>0 and xy=1.Find the minimum of the expression S=\(\left(x+y+1\right)\left(x^2+y^2\right)+\dfrac{4}{x+y}\)

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Kaya Renger Coordinator 28/12/2017 at 14:38

We have :

\(S=\left(x+y+1\right)\left(x^2+y^2\right)+\dfrac{4}{x+y}=\left(x+y\right)\left(x^2+y^2\right)+\left(x^2+y^2\right)+\dfrac{4}{x+y}\)

Applying Cauchy's inequality for two positive number , we have

\(S\ge2.\sqrt{\left(x+y\right)\left(x^2+y^2\right).\dfrac{4}{x+y}}+x^2+y^2=2.\sqrt{4\left(x^2+y^2\right)}+\left(x^2+y^2\right)\)

\(=4.\sqrt{\left(x^2+y^2\right)}+\left(x^2+y^2\right)=\sqrt{x^2+y^2}.\left(\sqrt{x^2+y^2}+5\right)\)

\(\ge1.\left(1+4\right)=5\)

So Min_S = 5 <=> x = y = 1
Selected by MathYouLike
Faded 22/01/2018 at 12:40

We have :

$S = (x + y + 1) (x^{2} + y^{2}) + \frac{4}{x + y} = (x + y) (x^{2} + y^{2}) + (x^{2} + y^{2}) + \frac{4}{x + y}$

Applying Cauchy's inequality for two positive number , we have

$S \geq 2. \sqrt{(x + y) (x^{2} + y^{2}) . \frac{4}{x + y}} + x^{2} + y^{2} = 2. \sqrt{4 (x^{2} + y^{2})} + (x^{2} + y^{2})$

$= 4. \sqrt{(x^{2} + y^{2})} + (x^{2} + y^{2}) = \sqrt{x^{2} + y^{2}} . (\sqrt{x^{2} + y^{2}} + 5)$

$\geq 1. (1 + 4) = 5$

So Min_S = 5 <=> x = y = 1

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