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sqing

#1 h Apr 14, 2025, 2:43 PM

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Let $x,y$ be reals such that $x+y=3$ and $\frac{1}{x^2+y}+\frac{1}{x+y^2}=\frac{1}{2}$ . Prove that
$x^2+y^2=7$ $x^3+y^3=18$ $x^4+y^4=47$

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Primeniyazidayi

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Primeniyazidayi

#2 h Apr 14, 2025, 3:11 PM

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Because $(x+y)^2 \geq 4xy$ ,it is clear that $xy \leq \frac{9}{4}$ .
Expanding the second equation gives $2(x^2+y^2) +6=x^3+y^3+xy(xy+1)=3(x^2-xy+y^2)+x^2y^2+xy$ or $x^2y^2+x^2+y^2-2xy=6$ or $x^2y^2-4xy+3=0$ .Factoring gives that $(xy-3)(xy-1)=0$ .Because of the first inequality,we have $xy=1$ and putting to equation gives $x^2+y^2=7$ .Then $x^3+y^3=3(x^2-xy+y^2)=18$ or just $x^3+y^3=18$ .Finally,because $x^4+y^4=(x^2+y^2)^2-2x^2y^2$ we have $x^4+y^4=47$ .Thus we are done.

This post has been edited 5 times. Last edited by Primeniyazidayi, Apr 14, 2025, 3:13 PM

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