# Proving conditions from inequality

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 $$p=q=0$$?

• "<<1" means almost zero – Rishi Sep 19 at 5:46

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 $$q=0$$.
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.