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inequality related to square the sum of any two sides of a triangle with respect to square of other side

Fastest way to check whether the triangle inequality is satisfiedThree sides of a $triangle$ are known. If a circle with it's center on base of $triangle$ touches the other two sides , find the radius of circle.Probability distribution of the third side in triangletriangle inequality theoremGeometry: Determining the length of a side of a triangleTriangle: One side, opposite angle and ratio of other sides givenObtuse Triangle geometry.A conjecture related to any triangleA conjecture about the sum of the areas of three triangles built on the sides of any given triangleProof for area of an equilateral triangle with respect to one side?

1

$begingroup$

Consider a,b,c are three side of a triangle. Now we need to find the relation between square the them sum of any two side of triangle with respect to third side.

My approach

As we know that the sum of any two sides of triangle will always be greater then the third side i.e

$(a+b)>c $

$(b+c)>a$

$(a+c)>b$

Take square on both side then

$a^2+b^2+2ab>c^2$

$ c^2+b^2+2bc>a^2$

$a^2+c^2+2ac>b^2$

Add last three equations

$2a^2+2b^2+2c^2-a^2-b^2-c^2>-2(ab+bc+ac)$

$a^2+b^2+c^2>-2(ab+bc+ac)$

Edited :
Corrected equation
But the correct answer is this $a^2 + b^2 + c^2 < 2(ab + bc + ca)$

What am I doing wrong?Please guide.

geometry triangles

share|cite|improve this question

edited Mar 20 at 18:20

Mostafa Ayaz

18.1k31040

asked Mar 20 at 18:00

ThinkerThinker

356

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Why not try your approach with inequalities like $a-c>-b$? This is just another form of your initial inequality.
  $endgroup$
  – Michael Burr
  Mar 20 at 18:06
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  But what do you want to prove? That $a^2+b^2+c^2geq2(ab+bc+ac)$?
  $endgroup$
  – Dr. Mathva
  Mar 20 at 18:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Dr.Mathva I want to prove this a2+b2+c2>2(ab+bc+ac)
  $endgroup$
  – Thinker
  Mar 20 at 18:09
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @MichaelBurr I will try this approach as well, but what is wrong with my approch?
  $endgroup$
  – Thinker
  Mar 20 at 18:10
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  But, @Thinker, the inequality $2(ab+bc+ac)geq a^2+b^2+c^2$ you want to prove is trivial from your last step
  $endgroup$
  – Dr. Mathva
  Mar 20 at 18:11
  
  
  

  
  
  
  

 | 
show 1 more comment

1

$begingroup$

Consider a,b,c are three side of a triangle. Now we need to find the relation between square the them sum of any two side of triangle with respect to third side.

My approach

As we know that the sum of any two sides of triangle will always be greater then the third side i.e

$(a+b)>c $

$(b+c)>a$

$(a+c)>b$

Take square on both side then

$a^2+b^2+2ab>c^2$

$ c^2+b^2+2bc>a^2$

$a^2+c^2+2ac>b^2$

Add last three equations

$2a^2+2b^2+2c^2-a^2-b^2-c^2>-2(ab+bc+ac)$

$a^2+b^2+c^2>-2(ab+bc+ac)$

Edited :
Corrected equation
But the correct answer is this $a^2 + b^2 + c^2 < 2(ab + bc + ca)$

What am I doing wrong?Please guide.

geometry triangles

share|cite|improve this question

edited Mar 20 at 18:20

Mostafa Ayaz

18.1k31040

asked Mar 20 at 18:00

ThinkerThinker

356

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Why not try your approach with inequalities like $a-c>-b$? This is just another form of your initial inequality.
  $endgroup$
  – Michael Burr
  Mar 20 at 18:06
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  But what do you want to prove? That $a^2+b^2+c^2geq2(ab+bc+ac)$?
  $endgroup$
  – Dr. Mathva
  Mar 20 at 18:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Dr.Mathva I want to prove this a2+b2+c2>2(ab+bc+ac)
  $endgroup$
  – Thinker
  Mar 20 at 18:09
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @MichaelBurr I will try this approach as well, but what is wrong with my approch?
  $endgroup$
  – Thinker
  Mar 20 at 18:10
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  But, @Thinker, the inequality $2(ab+bc+ac)geq a^2+b^2+c^2$ you want to prove is trivial from your last step
  $endgroup$
  – Dr. Mathva
  Mar 20 at 18:11
  
  
  

  
  
  
  

 | 
show 1 more comment

$begingroup$

Consider a,b,c are three side of a triangle. Now we need to find the relation between square the them sum of any two side of triangle with respect to third side.

My approach

As we know that the sum of any two sides of triangle will always be greater then the third side i.e

$(a+b)>c $

$(b+c)>a$

$(a+c)>b$

Take square on both side then

$a^2+b^2+2ab>c^2$

$ c^2+b^2+2bc>a^2$

$a^2+c^2+2ac>b^2$

Add last three equations

$2a^2+2b^2+2c^2-a^2-b^2-c^2>-2(ab+bc+ac)$

$a^2+b^2+c^2>-2(ab+bc+ac)$

Edited :
Corrected equation
But the correct answer is this $a^2 + b^2 + c^2 < 2(ab + bc + ca)$

What am I doing wrong?Please guide.

geometry triangles

share|cite|improve this question

edited Mar 20 at 18:20

Mostafa Ayaz

18.1k31040

asked Mar 20 at 18:00

ThinkerThinker

356

$endgroup$

share|cite|improve this question

edited Mar 20 at 18:20

Mostafa Ayaz

18.1k31040

asked Mar 20 at 18:00

ThinkerThinker

356

share|cite|improve this question

edited Mar 20 at 18:20

Mostafa Ayaz

18.1k31040

asked Mar 20 at 18:00

ThinkerThinker

356

edited Mar 20 at 18:20

Mostafa Ayaz

18.1k31040

edited Mar 20 at 18:20

Mostafa Ayaz

18.1k31040

asked Mar 20 at 18:00

ThinkerThinker

356

asked Mar 20 at 18:00

ThinkerThinker

356

2 Answers
2

active

oldest

votes

2

$begingroup$

Repeat your reasoning starting with this form of triangular inequalities:
$$
|a-b|<c,quad |b-c|<a,quad |c-a|<b,
$$
that is:
$$
(a-b)^2<c^2,quad (b-c)^2<a^2,quad (c-a)^2<b^2.
$$

share|cite|improve this answer

edited Mar 20 at 18:23

answered Mar 20 at 18:16

AretinoAretino

25.8k31545

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes this triangular inequalities is providing the correct result. But I still have one doubt left . What is wrong with my approach. triangle holds a property that sum of any two side is always greater then third side. Why reasoning based on this property is giving unexpected result?
  $endgroup$
  – Thinker
  Mar 20 at 18:28
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Your approach is fine, but leads to a trivial result, as you have on the right hand side a negative quantity and on the left hand side a positive quantity.
  $endgroup$
  – Aretino
  Mar 20 at 18:38
  

  
  
  
  

add a comment | 

1

$begingroup$

The inequality you attained is trivial as the lengths of sides of a triangle can't be negative or zero. In fact the correct answer comes from$$(a-b)^2+(a-c)^2+(b-c)^2< c^2+b^2+a^2$$which is a direct outcome of triangle inequality you mentioned above (how?).

share|cite|improve this answer

edited Mar 20 at 18:38

answered Mar 20 at 18:24

Mostafa AyazMostafa Ayaz

18.1k31040

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  My reasoning was based on the property that "The sum of the lengths of any two sides of a triangle is greater than the length of the third side. Similarly, the difference between the lengths of any two sides of a triangle is less than the length of the third side". But I understand that it is proving something that is not possible in as a,b, all will be greated than zero. But question is why?
  $endgroup$
  – Thinker
  Mar 20 at 18:36
  

  
  
  
  

add a comment | 

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2

$begingroup$

Repeat your reasoning starting with this form of triangular inequalities:
$$
|a-b|<c,quad |b-c|<a,quad |c-a|<b,
$$
that is:
$$
(a-b)^2<c^2,quad (b-c)^2<a^2,quad (c-a)^2<b^2.
$$

share|cite|improve this answer

edited Mar 20 at 18:23

answered Mar 20 at 18:16

AretinoAretino

25.8k31545

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes this triangular inequalities is providing the correct result. But I still have one doubt left . What is wrong with my approach. triangle holds a property that sum of any two side is always greater then third side. Why reasoning based on this property is giving unexpected result?
  $endgroup$
  – Thinker
  Mar 20 at 18:28
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Your approach is fine, but leads to a trivial result, as you have on the right hand side a negative quantity and on the left hand side a positive quantity.
  $endgroup$
  – Aretino
  Mar 20 at 18:38
  

  
  
  
  

add a comment | 

2

$begingroup$

Repeat your reasoning starting with this form of triangular inequalities:
$$
|a-b|<c,quad |b-c|<a,quad |c-a|<b,
$$
that is:
$$
(a-b)^2<c^2,quad (b-c)^2<a^2,quad (c-a)^2<b^2.
$$

share|cite|improve this answer

edited Mar 20 at 18:23

answered Mar 20 at 18:16

AretinoAretino

25.8k31545

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes this triangular inequalities is providing the correct result. But I still have one doubt left . What is wrong with my approach. triangle holds a property that sum of any two side is always greater then third side. Why reasoning based on this property is giving unexpected result?
  $endgroup$
  – Thinker
  Mar 20 at 18:28
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Your approach is fine, but leads to a trivial result, as you have on the right hand side a negative quantity and on the left hand side a positive quantity.
  $endgroup$
  – Aretino
  Mar 20 at 18:38
  

  
  
  
  

add a comment | 

$begingroup$

Repeat your reasoning starting with this form of triangular inequalities:
$$
|a-b|<c,quad |b-c|<a,quad |c-a|<b,
$$
that is:
$$
(a-b)^2<c^2,quad (b-c)^2<a^2,quad (c-a)^2<b^2.
$$

share|cite|improve this answer

edited Mar 20 at 18:23

answered Mar 20 at 18:16

AretinoAretino

25.8k31545

$endgroup$

share|cite|improve this answer

edited Mar 20 at 18:23

answered Mar 20 at 18:16

AretinoAretino

25.8k31545

answered Mar 20 at 18:16

AretinoAretino

25.8k31545

answered Mar 20 at 18:16

AretinoAretino

25.8k31545

1

$begingroup$

The inequality you attained is trivial as the lengths of sides of a triangle can't be negative or zero. In fact the correct answer comes from$$(a-b)^2+(a-c)^2+(b-c)^2< c^2+b^2+a^2$$which is a direct outcome of triangle inequality you mentioned above (how?).

share|cite|improve this answer

edited Mar 20 at 18:38

answered Mar 20 at 18:24

Mostafa AyazMostafa Ayaz

18.1k31040

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  My reasoning was based on the property that "The sum of the lengths of any two sides of a triangle is greater than the length of the third side. Similarly, the difference between the lengths of any two sides of a triangle is less than the length of the third side". But I understand that it is proving something that is not possible in as a,b, all will be greated than zero. But question is why?
  $endgroup$
  – Thinker
  Mar 20 at 18:36
  

  
  
  
  

add a comment | 

1

$begingroup$

The inequality you attained is trivial as the lengths of sides of a triangle can't be negative or zero. In fact the correct answer comes from$$(a-b)^2+(a-c)^2+(b-c)^2< c^2+b^2+a^2$$which is a direct outcome of triangle inequality you mentioned above (how?).

share|cite|improve this answer

edited Mar 20 at 18:38

answered Mar 20 at 18:24

Mostafa AyazMostafa Ayaz

18.1k31040

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  My reasoning was based on the property that "The sum of the lengths of any two sides of a triangle is greater than the length of the third side. Similarly, the difference between the lengths of any two sides of a triangle is less than the length of the third side". But I understand that it is proving something that is not possible in as a,b, all will be greated than zero. But question is why?
  $endgroup$
  – Thinker
  Mar 20 at 18:36
  

  
  
  
  

add a comment | 

$begingroup$

The inequality you attained is trivial as the lengths of sides of a triangle can't be negative or zero. In fact the correct answer comes from$$(a-b)^2+(a-c)^2+(b-c)^2< c^2+b^2+a^2$$which is a direct outcome of triangle inequality you mentioned above (how?).

share|cite|improve this answer

edited Mar 20 at 18:38

answered Mar 20 at 18:24

Mostafa AyazMostafa Ayaz

18.1k31040

$endgroup$

share|cite|improve this answer

edited Mar 20 at 18:38

answered Mar 20 at 18:24

Mostafa AyazMostafa Ayaz

18.1k31040

answered Mar 20 at 18:24

Mostafa AyazMostafa Ayaz

18.1k31040

answered Mar 20 at 18:24

Mostafa AyazMostafa Ayaz

18.1k31040