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Calculus tools render the minimization of $$\frac{(1-p)^2}{p} + \frac{p^2}{1-p}$$ on the interval $(0,1/2]$ to be a trivial task. But given how much symmetry there is in this expression, I was curious if one could use olympiad-style manipulations (AM-GM, Cauchy Schwarz, et al) to minimize this, show it is decreasing, or to show that it is bounded below by 1.

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  • $\begingroup$ The squares alone tell me to use Cauchy-Schwarz, but I can't see how exactly without screwing it up. $\endgroup$ – Sean Roberson May 23 '18 at 6:11
  • $\begingroup$ I applied AM-GM directly on the expression to bound it below by $2 \sqrt{p(1-p)}$, but that is a pretty bad bound for $p$ smaller than $1/2$. $\endgroup$ – user369210 May 23 '18 at 6:13
  • $\begingroup$ There is no minimum of the function on (0, 1/2). There is a minimum on (0, 1/2] which occurs at the boundary 1/2. What am I missing? $\endgroup$ – Andreas May 23 '18 at 6:21
  • $\begingroup$ Infimum, fine, same thing $\endgroup$ – user369210 May 23 '18 at 6:27
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This is elementary:

$$\frac{(1-p)^2}p+\frac{p^2}{1-p}=\frac{1-3p+3p^2-p^3+p^3}{p(1-p)}=\frac1{p(1-p)}-3.$$

The minimum is achieved by the vertex of the parabola $p(1-p)$, i.e. $p=\dfrac12\to\dfrac1{\frac14}-3$, as can be shown by completing the square.

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By Rearrangement inequality assuming wlog $p\ge(1-p)$

$$\frac{(1-p)^2}{p} + \frac{p^2}{1-p}=(1-p)\frac{1-p}{p} + p\frac{p}{1-p}\ge (1-p)\frac{p}{1-p} + p\frac{1-p}{p}=p+1-p=1$$

with equality for

$$p=1-p \implies p=\frac12$$

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  • $\begingroup$ The problem statement actually implies that $p \leq 1 - p.$ But of course this method still works, just swap $p$ and $1 - p.$ $\endgroup$ – David K May 24 '18 at 1:23
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** I modified this to give a proof with elementary methods only, no inequality theorems needed**

Simple transformations lead to $$ \frac{(1-p)^2}{p} + \frac{p^2}{1-p} = - 3 + \frac{4}{1 - 4 (p-1/2)^2} $$ Now it is easy to see that the global minimum of $f(p)$ for $p \in (0 , \; 1/2]$ occurs at $p = 1/2$.

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  • $\begingroup$ You've only shown that it is a local minimum. $\endgroup$ – Kenny Lau May 23 '18 at 6:30
  • $\begingroup$ The "small $x$" part invokes calculus, which I am trying to avoid $\endgroup$ – user369210 May 23 '18 at 6:33
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$$\begin{array}{rcll} \dfrac{(1-p)^2}{p} + \dfrac{p^2}{1-p} &=& \dfrac {(1-p)^3 + p^3} {p (1-p)} \\ &=& \dfrac {[(1-p) + p] [(1-p)^2 - p(1-p) + p^2]} {p (1-p)} \\ &=& \dfrac {(1-p)^2 + p^2} {p (1-p)} - 1 \\ &\ge& \dfrac {\left[ \frac1{\sqrt2} (1-p) + \frac1{\sqrt2} p \right]^2} {p (1-p)} - 1 &\text{Cauchy-Schwarz} \\ &=& \dfrac 1 {2 p (1-p)} - 1 \\ &=& \dfrac 1 {0.5 - 2 (p-0.5)^2} - 1 \\ &\ge& \dfrac 1 {0.5} - 1 \\ &=& 1 \end{array}$$

Both $\ge$ has equality at $p = 0.5$.

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$$\dfrac{(1-p)^2}p+\dfrac{p^2}{1-p}=\dfrac{1-3p+3p^2}{p(1-p)}=K\text{(say)}$$

$$\implies p^2(3+K)-p(3+K)+1=0$$

As $p$ is real, the discriminant must be $\ge0$

$$0\le(K+3)^2-4(K+3)=(K+3)(K-1)\implies$$

either $K\ge$max$(1,-3)$

or $K\le$min$(1,-3)$

But observe that $K+3\ne0$

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Make the change: $a=1-p, b=p, p\in \left(0,\frac12\right]$.

Then: $a+b=1, ab\le \frac14,$ the equality occurs when $a=b=\frac12$.

Hence: $$\frac{(1-p)^2}{p} + \frac{p^2}{1-p}=\frac{a^2}{b} + \frac{b^2}{a}=\frac{a^3+b^3}{ab}=\frac{(a+b)((a+b)^3-3ab)}{ab}\ge \frac{1\cdot \left(1^3-\frac34\right)}{\frac14}=1.$$

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Let's utilize the symmetry:

$$p\equiv\frac 1 2 -q$$

$$1-p\equiv\frac 1 2 +q$$

Then

$$ p \in (0,1/2] \iff q \in [0,1/2)$$

$$\frac{(1-p)^2}{p} + \frac{p^2}{1-p} \equiv \frac{(\frac 1 2 +q)^2}{\frac 1 2 -q} + \frac{(\frac 1 2 -q)^2}{\frac 1 2 +q} \equiv \dots$$

(here be magic, try it)

$$\dots \equiv \frac{\frac 1 4 + 3q^2}{\frac 1 4 -q^2} \equiv - 3 + \frac{4}{1 - 4q^2}$$

Note this is the result already obtained in another answer, however manipulations with $p$ (there) are not as elegant as manipulations with $q$ (here).

It's now easy to tell $q=0$ maximizes the denominator which is always positive (in the domain), so the last fraction is minimized.

$$ q=0 \iff p=\frac 1 2 $$

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