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Consider the following

$$ xp+qy \geq 0$$ where $x, y, p, q $ are real numbers. Here p and q are fixed, but x and y can be any real number with the condition that $$ \sqrt {x^2 +y^2} << 1 $$

Can we prove from here that in order for the above conditions to hold we must have $p=q=0$?

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  • $\begingroup$ "<<1" means almost zero $\endgroup$ – Rishi Sep 19 at 5:46
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Suppose $xp+yq \geq 0$ whenver $x^{2}+y^{2} <1$. Put $y=0$ to see that $xp \geq 0$ whenever $x^{2}<1$. Put $x=\frac 1 2$ to get $p \geq 0$ and put $x=-\frac 1 2$ to get $p \leq 0$. Hence $p=0$. Similarly $q=0$.

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Let do this: $$x=0 \ \& \ y=-q/\sqrt {q^2+1} \Longrightarrow xp+qy=-q^2/(q^2+1) \leqslant 0 \Longrightarrow q=0$$ $$y=0 \ \& \ x=-p/\sqrt {p^2+1} \Longrightarrow xp+qy=-p^2/(p^2+1) \leqslant 0 \Longrightarrow p=0$$ So YES, both must be zero.

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