1. ## Miscellaneous

Show that 1/3 ≤ (x2-2x+4)/(x2+2x+4) ≤ 3 for all real values of x.

1/3 ≤ (x2-2x+4)/(x2+2x+4) ≤ 3
let y=(x2-2x+4)/(x2+2x+4)
yx2+2yx+4y=x2-2x+4
(y-1)x2+(2y+2)x+4y-4=0
if x is real, then b2-4ac ≥ 0
(2y-2)2-4(y-1)(4y-4)≥0
y2-2y+1≤0
(y-1)2≤0

any idea???

2. ## Re: Miscellaneous

First of all, your proposed proof assumes what is to be proved. That kind of assumption can sometimes let you find a proof, but it can never be in a proof.

There is a fairly simple proof of this proposition if you know calculus.

$f(x) = \dfrac{x^2 - 2x + 4}{x^2 + 2x + 4} = 1 - \dfrac{4x}{x^2 + 2x + 4}.$

$\therefore f'(x) = -\ \dfrac{4(x^2 + 2x + 4) - (4x)(2x + 2)}{(x^2 + 2x + 4)^2} = \dfrac{-(4x^2 + 8x + 16) + (8x^2 + 8x)}{(x^2 + 2x + 4)^2} = \dfrac{4x^2 - 16}{(x^2 + 2x + 4)^2} \implies f'(x) = 0\ if\ x = \pm 2.$

$Also\ \displaystyle \lim_{x \rightarrow \infty}f(x) = 1 = \lim_{x \rightarrow - \infty}f(x).$

$f(2) = \dfrac{2^2 - 2(2) + 4}{2^2 + 2(2) + 4} \dfrac{4}{12} = \dfrac{1}{3}.$

$f(-2) = \dfrac{(-2)^2 - 2(-2) + 4}{(-2)^2 + 2(-2) + 4} \dfrac{12}{4} = 3.$

$\therefore \dfrac{1}{3} \le \dfrac{x^2 - 2x + 4}{x^2 + 2x + 4} \le 3.$

How you prove this without calculus is not readily apparent to me.

3. ## Re: Miscellaneous

Originally Posted by JeffM

How you prove this without calculus is not readily apparent to me.
This has frustrated me for days. It seemed to me that there SHOULD be a simple proof using algebra alone. There is. I tried to post it yesterday, but seem to have failed in doing so. Let's give it a second shot.

First we need this lemma:

$(x + 1)^2 \ge 0 \implies x^2 + 2x + 1 \ge 0 \implies (x^2 + 2x + 1) + 3 \ge 0 + 3 \implies x^2 + 2x + 4 \ge 3 \implies x^2 + 2x + 4 > 0.$

$Let\ f(x) = \dfrac{4x}{x^2 + 2x + 4}.$

$1 - f(x) = \dfrac{1}{1} - \dfrac{4x}{x^2 + 2x + 4} = \dfrac{x^2 + 2x + 4}{x^2 + 2x + 4} - \dfrac{4x}{x^2 + 2x + 4} = \dfrac{x^2 - 2x + 4}{x^2 + 2x + 4}.$

$0 \le 2(x + 2)^2 \implies 0 \le 2x^2 + 8x + 8 \implies -\ 4x \le - 4x + 2x^2 + 4x + 8 \implies -\ 4x \le 2x^2 + 4x + 8 \implies$

$-\ 4x \le 2(x^2 + 2x + 4) \implies \dfrac{-\ 4x}{x^2 + 2x + 4} \le 2 \implies -\ f(x) \le 2.$

$0 \le 2(x - 2)^2 \implies 0 \le 2x^2 - 8x + 8 \implies 12x \le 12x + 2x^2 - 8x + 8 \implies 12x \le 2x^2 + 4x + 8 \implies$

$3 * 4x \le 2(x^2 + 2x + 4) \implies \dfrac{4x}{x^2 + 2x + 4} \le \dfrac{2}{3} \implies f(x) \le \dfrac{2}{3} \implies -\ \dfrac{2}{3} \le -\ f(x).$

$In\ short, -\ \dfrac{2}{3} \le -\ f(x) \le 2 \implies \dfrac{1}{3} \le 1 - f(x) \le 3.$

$But,\ from\ above,\ 1 - f(x) = \dfrac{x^2 - 2x + 4}{x^2 + 2x + 4}.$

$THUS,\ \dfrac{1}{3} \le \dfrac{x^2 - 2x + 4}{x^2 + 2x + 4} \le 3.$

4. ## Re: Miscellaneous

Hello, Trefoil2727!

$\displaystyle \text{Show that }\:\frac{1}{3} \;\le \;\frac{x^2-2x+4}{x^2+2x+4} \;\le\;3\;\text{ for all real values of }x.$

First note that: $\displaystyle x^2-2x+4$ and $\displaystyle x^2+2x+4$ are both always positive.
. . Their minimum points are $\displaystyle (1,3)$ and $\displaystyle (\text{-}1,3)$, respectively.

Consider the first two terms: .$\displaystyle \frac{1}{3} \;\gtrless \;\frac{x^2-2x+4}{x^2+2x+4}$ .[1]

Cross-multiply: .$\displaystyle x^2+2x+4 \;\gtrless\;3x^2 - 6x + 12$

. . . . $\displaystyle 0 \;\gtrless\;2x^2 - 8x + 8 \quad\Rightarrow\quad 0\;\gtrless\;x^2-4x+4 \quad\Rightarrow\quad 0 \;\gtrless\;(x-2)^2$

Since $\displaystyle 0 \;\le \;(x-2)^2$, [1] becomes: .$\displaystyle \frac{1}{3} \;\le\;\frac{x^2-2x+4}{x^2+2x+4}$

Consider the last two terms: .$\displaystyle \frac{x^2-2x+4}{x^2+2x+4} \;\gtrless\;3$ .[2]

Cross-multiply: .$\displaystyle x^2-2x+4 \;\gtrless\;3x^2 + 6x + 12$

. . . . $\displaystyle 0\;\gtrless\;2x^2 + 8x + 8 \quad\Rightarrow\quad 0 \;\gtrless\;x^2 + 4x + 4 \quad\Rightarrow\quad 0 \;\gtrless\;(x+2)^2$

Since $\displaystyle 0 \:\le\:(x+2)^2$, [2] becomes: .$\displaystyle \frac{x^2-2x+4}{x^2+2x+4} \;\le\;3$

Therefore: .$\displaystyle \frac{1}{3} \;\le\;\frac{x^2-2x+4}{x^2+2x+4} \;\le\;3$

5. ## Re: Miscellaneous

Thanks Soroban. More concise and elegant solution than mine. I'll have to remember that greater or less than trick. I wasted a whole lot of time on this.

6. ## Re: Miscellaneous

Soroban, I have never before seen this symbol: \displaystyle \begin{align*} \gtrless \end{align*}. What is it used for? (It seems contradictory to me to make a statement that a quantity is both less than and greater than some other quantity...)

7. ## Re: Miscellaneous

Originally Posted by Prove It
Soroban, I have never before seen this symbol: \displaystyle \begin{align*} \gtrless \end{align*}. What is it used for? (It seems contradictory to me to make a statement that a quantity is both less than and greater than some other quantity...)
Prove It,

I never saw it before either, but I immediately knew what it meant. In doing this kind of problem I have sometimes used O to stand for unknown order relationship. That was always inelegant because some letters represented unknown numbers, but another letter stood for an unknown relationship. So the symbol stands for less than or equal to or greater than and is a tautology rather than a contradiction.