Prove that if $r$ is a real number such that $|r-1|<1$, then $\frac4{r(4-r)}\ge1$

Question

This was a bonus question on our last test. I'm not sure how begin. Would it be correct to say that since $|r-1|>0$, $0<r-1<2$?

The question is as follows:

Prove that if r is a real number such that $|r-1|<1$, then $\frac4{r(4-r)}\ge1$.

Any suggestions or help explaining what approach I should use would be appreciated.

Answer 1

Working forward gets you nowhere $[*]$. So we work backwards (one thing to keep in mind is this looks a heck of a lot like AM-GM).

$\frac 4{r(4-r)} \ge 1 \iff 0 < \frac {r(4-r)}4 \le 1 \iff 0 < r(4-r) \le 4$.

Now the question is to be clever or not. $r(4-r) = 4r - r^2$ and if we subtract $4$ from both sides we get $-4 < -(r-2)^2 < 0$ and [$**$]... well, you can't teach cleverness, so I'm going to pretend I am too dull to have seen that.

I want to compare the two factors $r$ and $4-r$ to $4$ which is $2*2$ so I'm thinking can we compare $r$ and $4-r$ each to $2$. If $r = 2 \pm d$ for some $d$ then $4-r = 2 \pm d$ and we get $r(4-r) = (2 + d)(2-d) = 4 - d^2 \le 4$.

That seems to have done it... but we never used $|r-1|< 2$ and ... ah... we also have to show $0 < r(4-r)$. As $|r-1| < 1$ then $-1 < r-1< 1$ so $0< r < 2$ so $r >0; 4-r > 2$ so $0 < r(4-r)$.

At this point, I realize that if $r >0; 4-r > 0$ then by AM-GM that $\sqrt{r(4-r)} \le \frac {r + 4 - r}2 = 2$ so $r(4-r) \le 4$ and ... well... that's nice. But IMO a prove that feels right, is better than something that accepts AM-GM on blind faith. We need to remember why AM-GM is true.

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$[*]$

$|r - 1|< 1 \implies$

$-1 < r -1 < 1 \implies $

$0 < r < 2 \implies $

$\frac 1r > \frac 12$. So? ... okay.....

And $0 < r < 2 \implies$

$4 - 0 > 4-r > 4 - 2 \implies$

$4 > 4-r > 2 \implies $

$\frac 14 < \frac 1{4-r} < \frac 12$. So? .... okay....

So $\frac 4{r(r-4)} > \frac 4{2(r-4)} > \frac 4{2*4} = \frac 12$ and .. ah, crap.

[$**$]

$ 0 < r(4-r) \le 4 \iff $

$0 < -r^2 + 4r \le 4 \iff $

$-4 < -(r^2 - 4r + 4) \le 0 \iff$

$0 \le (r-2)^2 < 4$.

And $(r-2)^2 \ge 0$ as all squares are. And $|r-1|< 1 \implies -1 < r-1 < 1 \implies -2 < r-2 < 0 \implies 0< (r-2)^2 < 4$.

Hmm...

Notice that $\frac 4{r(r-4)} \ne 1$ (unless $r = 2$). So $\frac 4{r(r-4)} > 1$ is a stronger result we could have proven.

Answer 2

HINT

Note that

$$|r-1|<1\iff -1<r-1<1 \iff0<r<2$$

and

• for $r(4-r)>0 \iff 0<r<4$

$$\frac{4}{r(4-r)}\ge1 \iff r(4-r)\le4 \iff r^2-4r+4\ge 0 \iff (r-2)^2\ge 0$$

• for $r(4-r)<0 \iff r<0 \quad\lor \quad r>4$

Answer 3

You are trying to prove $$r(4-r)-4<=0$$ which is $$r^2-4r+4>=0$$ then is equivalent to $${(r-2)}^2>=0$$ Is this clear enough?

Btw, of course, r cannot equal to 0 or 4 from the original question.