Minimum value of $p^2+q$ Let $p$ and $q$ be real numbers such that $p^2+q&gt;q^2+p$. Then the minimum value of $p^2+q$ is what?

My argument is as follows:

Since $p,q$ are real numbers, it should not matter if we take $p$ as $x$ and $q$ as $y$.
\begin{align*} p^2+q &amp;&gt; q^2+p \\ x^2-y^2 &amp;&gt; x-y \\ (x-y)(x+y-1)&amp;&gt;0\end{align*}
So, we have to maximize $x^2+y$ subject to the conditions that:
$$x-y&gt;0,\quad x+y-1&gt;0$$ or $$x-y&lt;0,\quad x+y-1&lt;0$$
From the graph of thse, it's clear that the minimum value occurs at $(1/2,1/2)$ in region $1$ (labeled on the graph). So, $p^2+q = (1/2)^2+(1/2) = 3/4$.


However this is not the correct answer.
Can anyone suggest where I'm making a mistake?
I'll solve the following problem.

Let $p^2+q\geq q^2+p$. Find a minimal value of $p^2+q$.

Indeed, $p^2+q+\frac{1}{4}=\frac{1}{2}\left(\left(p+\frac{1}{2}\right)^2+\left(q+\frac{1}{2}\right)^2\right)+\frac{1}{2}(p^2+q-q^2-p)\geq0$.

The equality occurs for $p=q=-\frac{1}{2}$.

Id est, the answer is $-\frac{1}{4}$.
It appears that you made an error early on in your proof. You got (x-y)(x+y-1)>0 when you should've gotten x+y-1>0, as the x-y terms cancel each other out.