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 November 12th, 2011, 02:43 AM #1 Newbie   Joined: Nov 2011 Posts: 2 Thanks: 0 to prove inequality Suppose x > -1, then x/(x+1) < (x+1)/(x+2) this needs to be proven ^ I tried to go direct and I ended up with this x(x+1)(x+2)^2 < (x+1)^3(x+2) I don't know where to go to from there o.O I tried using induction to continue but it doesn't work... Thoughts?
 November 12th, 2011, 05:52 AM #2 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,878 Thanks: 1087 Math Focus: Elementary mathematics and beyond Re: to prove inequality x/(x + 1) < (x + 1)/(x + 2) x/(x + 1) - (x + 1)/(x + 2) < 0 (x² + 2x - x² - 2x - 1)/((x + 1)(x + 2)) < 0 -1/((x + 1)(x + 2)) < 0. The denominator is a parabola that opens upward and has zeros -1 and -2 so the denominator is negative for -2 < x < -1, so the inequality x/(x + 1) < (x + 1)/(x + 2) is true for x > -1, x < -2.
 November 12th, 2011, 06:17 AM #3 Math Team   Joined: Apr 2010 Posts: 2,778 Thanks: 361 Re: to prove inequality Or $\frac{x}{x + 1} < \frac{x+1}{x + 2}$ For $(x+1)(x+2) > 0$, true for $x<-2$ and $x>-1$ $\frac{x}{x + 1}(x+1)(x+2) < \frac{x+1}{x + 2}(x+1)(x+2)$ $x(x+2) < (x+1)^2$ true for all x. Hence $x<-2$ and $x>-1$
November 12th, 2011, 08:07 AM   #4
Math Team

Joined: Dec 2006
From: Lexington, MA

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Re: to prove inequality

Hello, jamesb1!

Quote:
 $\text{Suppose }\,x\,>\, -1,\,\text{ then: }\:\frac{x}{x\,+\,1} \:< \:\frac{x\,+\,1}{x\,+\,2}$

$\text{Since }x\,>\,-1,\,\text{ then: }\:x\,+\,1 \:>\:0\,\text{ and }\,x\,+\,2\:>\:0$

$\text{W\!e begin with: }\:0 \;<\;1$

$\text{Add }x^2\,+\,2x\text{ to both sides:}$
[color=beige]. . [/color]$x^2\,+\,2x\;<\;x^2\,+\,2x\,+\,1$

$\text{And we have:}$
[color=beige]. . [/color]$x(x\,+\,2) \;<\;(x\,+\,1)^2$

$\text{Divide by }(x\,+\,1)(x\,+\,2),\,\text{ a positive quantity:}$

[color=beige]. . [/color]$\frac{x}{x\,+\,1} \;<\;\frac{x\,+\,1}{x\,+\,2}$

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