inequality related to square the sum of any two sides of a triangle with respect to square of other sideFastest 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?
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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?
$begingroup$
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
$endgroup$
|
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
$endgroup$
$begingroup$
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
$endgroup$
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
geometry triangles
edited Mar 20 at 18:20
Mostafa Ayaz
18.2k31040
18.2k31040
asked Mar 20 at 18:00
ThinkerThinker
355
355
$begingroup$
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
$begingroup$
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
$begingroup$
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
$begingroup$
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
$begingroup$
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
$begingroup$
@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
$begingroup$
@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
$begingroup$
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
2 Answers
2
active
oldest
votes
$begingroup$
Repeat your reasoning starting with this form of triangular inequalities:
$$
|a-b|<c,quad |b-c|<a,quad |c-a|<b,
$$
that is:
$$
(a-b)^2<c^2,quad (b-c)^2<a^2,quad (c-a)^2<b^2.
$$
$endgroup$
$begingroup$
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 |
$begingroup$
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?).
$endgroup$
$begingroup$
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 |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Repeat your reasoning starting with this form of triangular inequalities:
$$
|a-b|<c,quad |b-c|<a,quad |c-a|<b,
$$
that is:
$$
(a-b)^2<c^2,quad (b-c)^2<a^2,quad (c-a)^2<b^2.
$$
$endgroup$
$begingroup$
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:
$$
(a-b)^2<c^2,quad (b-c)^2<a^2,quad (c-a)^2<b^2.
$$
$endgroup$
$begingroup$
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:
$$
(a-b)^2<c^2,quad (b-c)^2<a^2,quad (c-a)^2<b^2.
$$
$endgroup$
Repeat your reasoning starting with this form of triangular inequalities:
$$
|a-b|<c,quad |b-c|<a,quad |c-a|<b,
$$
that is:
$$
(a-b)^2<c^2,quad (b-c)^2<a^2,quad (c-a)^2<b^2.
$$
edited Mar 20 at 18:23
answered Mar 20 at 18:16
AretinoAretino
25.8k31545
25.8k31545
$begingroup$
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 |
$begingroup$
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
$begingroup$
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
$begingroup$
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
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
$begingroup$
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 |
$begingroup$
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?).
$endgroup$
$begingroup$
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?).
$endgroup$
$begingroup$
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?).
$endgroup$
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?).
edited Mar 20 at 18:38
answered Mar 20 at 18:24
Mostafa AyazMostafa Ayaz
18.2k31040
18.2k31040
$begingroup$
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 |
$begingroup$
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
$begingroup$
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
$begingroup$
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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$begingroup$
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