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 October 18th, 2018, 12:38 PM #1 Senior Member     Joined: Oct 2016 From: Arizona Posts: 193 Thanks: 34 Math Focus: I'm still deciding, but my recent focus has been olympiad problems and math journal problems. Inequality tips Hi, so I'm working on an inequality problem, and the problem gives you that $a,b,c$ are real numbers and $a+b+c=3$. Using this and Cauchy Schwarz, I was able to get that $a^2+b^2+c^2 \ge 3$ and $-3 \le ab+bc+ca \le 3$. This helps out, but there is still $abc$ I would like to know something about, but can't seem to figure out any bounds on it. Does anyone have any ideas? Specifically there is a term $+18abc$ which I'd like to be $\ge 0$ and I've written that as $2(a+b+c)^2(abc)$ but we obviously can't conclude that $abc$ is \ge0 since they are real numbers. Sorry if this is confusing. I'm mainly looking for ideas on what I could do.
 October 18th, 2018, 01:25 PM #2 Senior Member   Joined: Sep 2016 From: USA Posts: 578 Thanks: 345 Math Focus: Dynamical systems, analytic function theory, numerics It seems you are leaving out some information. You can't conclude anything about the sign of $abc$ for exactly the reasons you have stated.
October 18th, 2018, 01:39 PM   #3
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Joined: Oct 2016
From: Arizona

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Math Focus: I'm still deciding, but my recent focus has been olympiad problems and math journal problems.
Quote:
 Originally Posted by SDK It seems you are leaving out some information. You can't conclude anything about the sign of $abc$ for exactly the reasons you have stated.
I can just figure it out, thanks!

 October 18th, 2018, 03:14 PM #4 Member   Joined: Oct 2018 From: Netherlands Posts: 39 Thanks: 3 There are no bounds to $abc$. For instance, with a large positive $bound$ take the following values: $a = +bound$ $b = -bound+4$ $c = -1$ The special case is where $a, b, c$ are all positive real numbers. Not hard to see that $0 <= abc <= 1$ Last edited by Arisktotle; October 18th, 2018 at 03:17 PM.
October 18th, 2018, 03:23 PM   #5
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Joined: Oct 2016
From: Arizona

Posts: 193
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Math Focus: I'm still deciding, but my recent focus has been olympiad problems and math journal problems.
Quote:
 Originally Posted by Arisktotle There are no bounds to $abc$. For instance, with a large positive $bound$ take the following values: $a = +bound$ $b = -bound+4$ $c = -1$ The special case is where $a, b, c$ are all positive real numbers. Not hard to see that $0 <= abc <= 1$
Okay, thank you! I will try to incorporate this term into the rest of the inequality. I appreciate it!

 October 19th, 2018, 10:40 AM #6 Senior Member     Joined: Oct 2016 From: Arizona Posts: 193 Thanks: 34 Math Focus: I'm still deciding, but my recent focus has been olympiad problems and math journal problems. This is my 5th inequality proof and it's crazy how many tricks of the trade I learned, I had to completely change my approach! Lesson: if an approach isn't working, don't keep trying it

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