Minimum value of $p^2+q$

Question

Let $p$ and $q$ be real numbers such that $p^2+q>q^2+p$. 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 &> q^2+p \\ x^2-y^2 &> x-y \\ (x-y)(x+y-1)&>0\end{align*} So, we have to maximize $x^2+y$ subject to the conditions that: $$x-y>0,\quad x+y-1>0$$ or $$x-y<0,\quad x+y-1<0$$ 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?

Answer 1

Given $p\in \mathbb{R}$, we find the set of $q\in\mathbb{R}$ that satisfy $p^2+q>q^2-p$. Rewrite the inequality:

$$q^2-q-(p^2+p)<0$$

The roots of $q^2-q-(p^2+p)=0$ are

$$q=\frac{1\pm\sqrt{1+4\left(p^2+p\right)}}{2}$$

Notice that $4p^2+4p+1={(2p+1)}^2$, and hence the roots are $p$ and $p+1$. Thus, $q \in (p,p+1)$.

For any fixed $p$, the task of minimizing $p^2+q$ therefore reduces to minimizing $q\in(p,p+1)$. We see the infimum is $p^2+p$, but there is no actual minimum.

Finally, letting $p$ (and thus $q$) vary throughout $\mathbb{R}$, the minimum of $p^2+p$ can be easily found to be $p=-\frac12$. Any methods for min-maxing a quadratic equation will do.

Answer 2

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.

Answer 3

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 $-\frac{1}{4}$.

Answer 4

Solving the problem with the following condition $p^2+q>=q^2+p$, since minimum won't exist for open set. Fist of all its clear that minimum won't lie in the region 1. Second the gradient of $x^2+y$ is $(2x,1)$ which will never vanish, which means that the minimum can not lie inside the region because you can always decrease y to get a better solution. So minimum can only lie on the boundaries, now it won't lie on $x+y=1$, as you can improve the solution by taking corresponding point on $x=y$(by corresponding I mean reflection through $y=1/2)$. So It can only lie on $x=y$, so what we have to do is find minimum of $x^2+x$, which will occur at $x=-1/2$. So the answer is -1/4

Answer 5

From what is given; $p^2+q \ge \frac{1}{2}(p^2+q+q^2+p) \ge \frac{-1}{4}$

The last step obviously can easily be explained using quadratic formula, AM-GM etc. or whatever you like.