15

Recently I've came across such problem: give a polynomial $P(x,y)$, with $\inf_{\mathbb{R}^2} P=0$, but there is no point on the plane where $P=0$. I couldn't solve it after a day, and seriously doubt whether such a function exists, however its source claims that there is. Is that really possible?

hardmath
  • 35,235
  • 19
  • 69
  • 133
aplavin
  • 591
  • 1
  • 4
  • 18

1 Answers1

28

$P(x,y)=(1-xy)^2+x^2$ has this property. Clearly $P>0$ and also the sequence $(x_n,y_n)=(1/n,n)$ shows that $\inf P=0$.

JP McCarthy
  • 8,008
  • 1
  • 31
  • 55