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I have three variables $p \geq 0$, $q \geq 0$, $r \geq 0$ and a positive constant $m$. Let $m = p + q +r$.

How can I show that the maximum value of $pq + r$ is no more than $\frac{m^2}{4}$?

It's easy to see that decreasing $r$ (thus increasing $p + q$) will give us higher value. It's also easy to see that the maximum value of $pq + r$ when $r = 0$ is achieved at $p = q = \frac{m}{2}$. But I believe there must be mathematically better way to proof this.

Arimelho
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  • There must be an error here. The constraint on $m$ is given because $p,q,r \geq 0$. Moreover, there should be no maximum since you can let $p,q,r$ be arbitrarily large. I don't see how that connects to $m$ in any way--even with $\frac{m^2}{4}=\frac{1}{4}(p+q+r)^2$. – mathematics2x2life Nov 12 '13 at 06:52
  • @mathematics2x2life Looks like I made a little mistake on the formulation. I updated the question to better describe the problem. – Arimelho Nov 12 '13 at 07:00

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$m=p+q+r$,

$r=m-p-q$.

So you want to maximize the : $pq+r=pq+m-p-q=m+(pq-p-q)$

Since m is stable, you need to maximize (pq-p-q). You can easily show that this is maximized when p=q.

So now you have to find the maximum value of: $f(p)=m+p^2-2p$, where p belongs in the $[0,\frac{m}2]$ (0 when r=m and $\frac{m}2$ when r=0)

In order to find the maximum, you need to examine

  • where the derivative is zero
  • the edges

f'(p)=2p-2, f''(p)=2>0. so, this is a minimum.

f(0)=m

$f(\frac{m}2)=\frac{m^2}4$

So your maximum is the biggest value of m and $\frac{m^2}4$ (and not just $\frac{m^2}4$ as you claimed. imagine p=0,q=0, r=1)

Thanos Darkadakis
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