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I've been working on this problem for a while and just can't quite seem to figure out what to do. I'm not so much looking for a solution as I am a push in the right direction. I was able to show that

$0<x(1-x)<1$

but this is where I started having trouble. We are not supposed to use calculus for this and, if possible, I would like to avoid using the formula for the maximum of a quadratic. I understand that if I can show that

$x(1-x)<\frac {1}{4}$

then it would be easy as I can then provide an instance where

$x(1-x)=\frac {1}{4}$

and thus

$x(1-x) \leq \frac {1}{4}$

Jacob Hulse
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  • $x(1-x)=(\frac12+(x-\frac12))\cdot(\frac12-(x-\frac12))=\frac14-(x-\frac12)^2\le\frac14$ – Hagen von Eitzen Feb 12 '20 at 19:51
  • You say not to use calculus. Do limits count as calculus here? – R. Burton Feb 12 '20 at 19:58
  • @MartinR Yes, I must have skimmed over that. Thanks. – Jacob Hulse Feb 12 '20 at 20:05
  • Hint: Let $x=\frac 12 + r$. That would mean $1-x= \frac 12 -r$. So you need to prov that $\frac 1{x(1-x)}= \frac 1{(\frac 12-r)(\frac 12 + r)}= \frac 1{\frac 14 - r^2} \ge 4$. Which as $0 < x(1-x) = (\frac 12-r)(\frac 12 +r) = \frac 14 -r^2 \le \frac 14$ is clear. – fleablood Feb 12 '20 at 20:16

4 Answers4

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Hint:

$x=\frac{1}{2}$ is the only value $x\in (0,1)$ such that $x(1-x)=\frac{1}{4}$

Emilio Novati
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If $0<x<1$, then $x(1-x)>0$. So the inequality you are looking at is equivalent to $$x(1-x)\geq \frac{1}{4}$$ Note that $x(1-x)$ is a parabola. The zeroes of this parabola are at $x=0$ and $x=1$, and its vertex (the "tip" of the curve) is right in the middle of them. What is its vertex?

the parabola x(1-x)

Luiz Cordeiro
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If $0<x<1$ then also $0<1-x<1$. So $$\frac{1}{x(1-x)}\ge4\iff 4x(1-x)\le 1 \iff 4x^2-4x+1\ge 0$$

Now, can you solve this last problem?

Tito Eliatron
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Using AM-GM:

$$\sqrt{x(1-x)} \leq \frac{x+1-x}{2}=\frac{1}{2}$$

Squaring and flipping the inequality gives:

$$\frac{1}{x(1-x)} \geq 4$$

LHF
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