Inequality of sides of triangle

[sharpycasio]

Homework Statement

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

-1<\frac{a}{b}+\frac{b}{c}+\frac{c}{a}-\frac{b}{a}-\frac{a}{c}-\frac{c}{b}<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<(\frac{a}{b}-\frac{b}{a})+(\frac{b}{c}-\frac{c}{b})+(\frac{c}{a}-\frac{a}{c})<1

I'm stuck. Any help? Thanks.

 

[sharpycasio]

Can anyone please give me a hint? Thanks.

 

[happysauce]

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.

 

[sharpycasio]

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.

 

[happysauce]

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?

 

[SammyS]

Homework Statement

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

-1<\frac{a}{b}+\frac{b}{c}+\frac{c}{a}-\frac{b}{a}-\frac{a}{c}-\frac{c}{b}<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<(\frac{a}{b}-\frac{b}{a})+(\frac{b}{c}-\frac{c}{b})+(\frac{c}{a}-\frac{a}{c})<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 />

 

[sharpycasio]

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 :)