What is the Triangle Inequality in an Acute Triangle?

In summary, the conversation is about trying to prove or disprove an inequality in any acute triangle ABC with sides a, b, and c. The inequality in question is 27 ≤ (a+b+c)^2 (1/a^2 + 1/b^2 - 1/c^2 + 1/b^2 + 1/c^2 - 1/a^2 + 1/c^2 + 1/a^2 - 1/b^2). The participants in the conversation offer advice and methods for solving the problem, such as using the cosine law, Heron's formula, and the Arithmetic-Geometric inequality. Ultimately, it is determined that the inequality is true and can be proven by showing that (
  • #1
Bitter
98
0
I am a beginner at trying to prove or disprove inequalities. In an attempt to improve on this skill I found some problems that I would like to work on. Now, I know many of you may be able to look at this and think of a solution, but please refrain from posting it, but some advice and methods would be helpful.

Prove that in any acute triangle ABC with sides a, b, and c, the following inequality is true. [tex] 27 \leq (a+b+c)^2 \left( \frac{1}{a^2 + b^2 - c^2}+\frac{1}{b^2 + c^2 - a^2}+\frac{1}{c^2 + a^2 - b^2} \right) [/tex]
 
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  • #2
Are you trying to say:

[tex] 27 \leq (a+b+c)^2 \left( \frac{1}{a^2 + b^2 - c^2}+\frac{1}{b^2 + c^2 - a^2}+\frac{1}{c^2 + a^2 - b^2} \right) [/tex]

what have you tried?
 
  • #3
Well, here is where I'm at. Well, I am fairly certain that this equation is true. If we exam the smallest acute triangle I can think of, an equilateral triangle whose sides are all one, then the solutions comes out to be exactly 27. So, that is where i am at right.
 
  • #4
And what makes you think the RHS will be smaller for acute triangles?

I don't immediately see a clever way to do this, so I would start by multiplying out by a common denominator so that you're comparing polynomials, and then simplifying as much as possible.
 
  • #5
mmm...problem like this is reminiscent of those you usuallly see in maths competitions or olympiad. It usually require some wit to do it elegantly. If you have absolutely no idea (like me currently :smile:), you should start by thinking about all the properties of an acute triangle and see whether that gives you other contraints relations between a, b, c. Next think about why it may not work for a triangle that is NOT acute (i guess that means it has one angle bigger than 90)

next, look at the expression itself (change its form) ...and see if you can recognise it or part of it from somewhere.. if not, see if you can find another inequality which is easier to handle that is bigger than or equal 27 yet less than or equal to original expression.

remember there are basic things like a + b > c, a + c> b, b + c > a. etc.
 
  • #6
Using the cosine law, you can find an inequality for which an angle is acute. They talk about it here: http://mathworld.wolfram.com/AcuteTriangle.html"
 
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  • #8
given the similarity of the three terms in brackets, may be 27 really means 9+9+9, and proving that each term is [tex]9 \leq[/tex] may be sufficient... just a guess.
Certainly, acute angle means that Cos (any angle) is inside the open interval [tex](0,1)[/tex]
 
  • #9
Let a=b a=c

27<_(a + a + a)^2((1/a^2) + (1/a^2) + (1/a^2))
27<_(9a^2)(3/a^2)
27<_27

So that proves that an acute triangle that happens to be an equilateral triangle is true, that means I just have to show that the rest hold true.
 
  • #10
Yet another trivial problem you couldn't tackle ,ha Bitter ?:biggrin:
 
  • #11
tehno said:
Yet another trivial problem you couldn't tackle ,ha Bitter ?:biggrin:

I'm not really sure what your problem is with me. I'm not even sure why you said yet again. The other problem I posted was not one I couldn't solve. I merely wanted to see other people's solutions to it. I just posted it in the wrong place and forgot about it. In fact, I'm fairly certain this is my first post about solving a problem in general math section.

So if you don't plan to give me advice that's fine, but if you feel the need to rub in the fact that you are better at math than I am, that's other thing.
 
  • #12
Bitter said:
I'm not really sure what your problem is with me. I'm not even sure why you said yet again. The other problem I posted was not one I couldn't solve. I merely wanted to see other people's solutions to it.
Really?I don't recall I've seen you ever posted your solution..But never mind that.It was another problem anyway.I got an impression your main problem isn't math or misplacing the posts,but lazyness.Again,that's just my opinion.This time you posted the problem to the right place and you will see me post the solution.But only this time.


[tex]S=\frac{1}{a^2+b^2-c^2}+\frac{1}{b^2+c^2-a^2}+\frac{1}{a^2+c^2-b^2}\geq \frac{27}{(a+b+c)^2}[/tex]

(Firstly note the right side of ineq. is always positive-> left side must be always positive too.That is ensured ,becouse [itex]a^2+b^2-c^2>0,b^2+c^2-a^2>0,a^2+c^2-b^2>0[/itex]
is always true for acute triangles.Hence ,the restriction)

We have:

[tex]\frac{a^2}{a^2b^2+a^2c^2-a^4}+\frac{b^2}{a^2b^2+b^2c^2-b^4}+\frac{c^2}{a^2c^2+b^2c^2-c^4}=S \geq \frac{(a+b+c)^2}{2(a^2b^2+a^2c^2+b^2c^2)-(a^4+b^4+c^4)}=\frac{a+b+c}{(a+b-c)(a+c-b)(b+c-a)}[/tex]

Now we are left to prove that for a,b,c>0 the following is true:

[tex]\frac{a+b+c}{(a+b-c)(a+c-b)(b+c-a)}\geq \frac{27}{(a+b+c)^2}[/tex]


It's true becouse ,by Arithmetic-Geometric ineq,we have [itex](a+b+c)^3\geq 27abc[/itex] while [itex]abc\geq (a+b-c)(a+c-b)(b+c-a)[/itex] is obvious.
QED

EDIT:I can't be better mathematician than you.That's becouse I'm not a mathematician (per education) at all.
:smile:

Best regards
 
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  • #13
Sigh, I thought I said not to post an answer, but to help me out. Oh well...
 

What is the Triangle Inequality?

The Triangle Inequality is a fundamental geometric concept that states that the sum of the lengths of any two sides of a triangle must be greater than the length of the third side.

Why is the Triangle Inequality important?

The Triangle Inequality is important because it helps us determine if a given set of three side lengths can form a valid triangle. It also has many applications in mathematics and physics, such as in the study of vectors and inequalities.

How is the Triangle Inequality used in geometry?

In geometry, the Triangle Inequality is used to prove the properties and relationships of triangles. It is also used to determine if a given set of side lengths can form a triangle, and to find the range of possible values for the third side of a triangle.

What is the difference between the Triangle Inequality and the Triangle Equality?

The Triangle Inequality states that the sum of any two sides of a triangle must be greater than the length of the third side, while the Triangle Equality states that the sum of any two sides must be equal to the length of the third side. The Triangle Inequality is a strict inequality, while the Triangle Equality is an equality.

How can the Triangle Inequality be extended to other shapes?

The Triangle Inequality can be extended to other shapes by applying the concept of side lengths and their relationships. For example, in a quadrilateral, the sum of any three sides must be greater than the length of the fourth side, and in a regular polygon, the sum of any two sides must be greater than the length of the remaining side.

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