Inequality of sides of triangle

AI Thread Summary
The discussion centers on proving the inequality for the sides of a triangle: -1 < (a/b + b/c + c/a - b/a - a/c - c/b) < 1. Participants suggest using the Law of Sines or Cosines and explore rearranging the terms to find a solution. One user proposes assuming a > b > c to simplify the problem, while another mentions expanding the expression and factoring the numerator. Ultimately, the problem is resolved by applying the triangle inequality to demonstrate the validity of the inequality. The conversation highlights the importance of foundational trigonometric concepts in solving geometric inequalities.
sharpycasio
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Homework Statement


Prove the following inequality for any triangle that has sides a, b, and c.

-1&lt;\frac{a}{b}+\frac{b}{c}+\frac{c}{a}-\frac{b}{a}-\frac{a}{c}-\frac{c}{b}&lt;1

Homework Equations


The Attempt at a Solution



I think we have to use sine or cosine at a certain point because the bounds of the inequality are the same as the bounds of the two functions' ranges. Perhaps the Sine Law since that applies to all triangles? Tried rearranging it, pairing up the reciprocals. Maybe the fractions represent ratios (sin(\theta))

-1&lt;(\frac{a}{b}-\frac{b}{a})+(\frac{b}{c}-\frac{c}{b})+(\frac{c}{a}-\frac{a}{c})&lt;1

I'm stuck. Any help? Thanks.
 
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Can anyone please give me a hint? Thanks.
 
sharpycasio said:
Can anyone please give me a hint? Thanks.

What have you covered in the class, there are many many many trig and geometry topics, maybe showing some theorems might narrow the search down.
 
happysauce said:
What have you covered in the class, there are many many many trig and geometry topics, maybe showing some theorems might narrow the search down.

I've done all the trig usually done at the high school level. My analytic geometry skills aren't that good though.

Can someone please help me? I need to solve this for tomorrow. I've been working on it for like 5 hours and I'm still stuck.
 
Well I took a crack at it. Here is what I tried. No guarantee this will work.
Without loss of generality let a>b>c. With equality it's pretty clear that is works. I tried playing with this assumption, not sure if it will work though. Another idea would be trying to get something to take the form of law of cosines and sines. Maybe you should try assuimg a>b>c, then use the idea that a+b>c, b+c>a, a+c>b. To be honest, I think there is something really important that I missed. Maybe if you list every theorem you learned in the past 2 weeks?
 
sharpycasio said:

Homework Statement


Prove the following inequality for any triangle that has sides a, b, and c.

-1&lt;\frac{a}{b}+\frac{b}{c}+\frac{c}{a}-\frac{b}{a}-\frac{a}{c}-\frac{c}{b}&lt;1

Homework Equations


The Attempt at a Solution



I think we have to use sine or cosine at a certain point because the bounds of the inequality are the same as the bounds of the two functions' ranges. Perhaps the Sine Law since that applies to all triangles? Tried rearranging it, pairing up the reciprocals. Maybe the fractions represent ratios (sin(\theta))
-1&lt;(\frac{a}{b}-\frac{b}{a})+(\frac{b}{c}-\frac{c}{b})+(\frac{c}{a}-\frac{a}{c})&lt;1
I'm stuck. Any help? Thanks.
I don't have a final answer but here are some thoughts:

Law of sines will give you: \displaystyle \frac{a}{b}=\frac{\sin(\text{A}}{\sin(\text{B}}

\displaystyle <br /> \frac{\sin(\text{A})}{\sin(\text{B})}-\frac{\sin(\text{B})}{\sin(\text{A})}<br /> =\frac{\sin^2(\text{A})-\sin^2(\text{B})}{\sin(\text{A})\,\sin(\text{B})}<br /> =\frac{\cos(\text{A}+\text{B})}{\sin(\text{A})\, \sin(\text{B})}<br />
 
SammyS said:
I don't have a final answer but here are some thoughts:

Law of sines will give you: \displaystyle \frac{a}{b}=\frac{\sin(\text{A}}{\sin(\text{B}}

\displaystyle <br /> \frac{\sin(\text{A})}{\sin(\text{B})}-\frac{\sin(\text{B})}{\sin(\text{A})}<br /> =\frac{\sin^2(\text{A})-\sin^2(\text{B})}{\sin(\text{A})\,\sin(\text{B})}<br /> =\frac{\cos(\text{A}+\text{B})}{\sin(\text{A})\, \sin(\text{B})}<br />

This is exactly how I started working on it but then I got stuck. I ended up solving it a different way. I expanded the whole expression on a common denominator (abc). Then I factored the numerator into −(a−b)(a−c)(b−c) (See https://www.physicsforums.com/showthread.php?t=643010)

Then I used the triangle inequality as happysauce said. a+b>c → a-c>-b and "substituted". The rest was pretty simple. Thanks for the help guys :)
 
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