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Help me proving the inequality [on hold]

-2

$begingroup$

If $a,b,c$ are sides of a triangle then prove that,
$$left(fraca+bcright)^3+left(fracb+caright)^3+left(fracc+abright)^3 < 24$$

inequality

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edited Mar 28 at 10:06

Sujit Bhattacharyya

1,632519

asked Mar 28 at 9:55

Avishek MitraAvishek Mitra

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put on hold as off-topic by Martin R, Javi, Delta-u, YiFan, Adrian Keister Mar 28 at 16:56

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Martin R, Javi, Delta-u, YiFan, Adrian Keister

If this question can be reworded to fit the rules in the help center, please edit the question.

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  Hello, and welcome to math.se! Please add a blurb of what you've tried :) Also, please use the latex syntax to format your question.
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  – Siddharth Bhat
  Mar 28 at 9:57
  

  
  
  
  

add a comment | 

-2

$begingroup$

If $a,b,c$ are sides of a triangle then prove that,
$$left(fraca+bcright)^3+left(fracb+caright)^3+left(fracc+abright)^3 < 24$$

inequality

share|cite|improve this question

edited Mar 28 at 10:06

Sujit Bhattacharyya

1,632519

asked Mar 28 at 9:55

Avishek MitraAvishek Mitra

6

New contributor

Avishek Mitra is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$

put on hold as off-topic by Martin R, Javi, Delta-u, YiFan, Adrian Keister Mar 28 at 16:56

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Martin R, Javi, Delta-u, YiFan, Adrian Keister

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Hello, and welcome to math.se! Please add a blurb of what you've tried :) Also, please use the latex syntax to format your question.
  $endgroup$
  – Siddharth Bhat
  Mar 28 at 9:57
  

  
  
  
  

add a comment | 

$begingroup$

If $a,b,c$ are sides of a triangle then prove that,
$$left(fraca+bcright)^3+left(fracb+caright)^3+left(fracc+abright)^3 < 24$$

inequality

share|cite|improve this question

edited Mar 28 at 10:06

Sujit Bhattacharyya

1,632519

asked Mar 28 at 9:55

Avishek MitraAvishek Mitra

6

New contributor

Avishek Mitra is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this question

edited Mar 28 at 10:06

Sujit Bhattacharyya

1,632519

asked Mar 28 at 9:55

Avishek MitraAvishek Mitra

6

New contributor

Avishek Mitra is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

edited Mar 28 at 10:06

Sujit Bhattacharyya

1,632519

asked Mar 28 at 9:55

Avishek MitraAvishek Mitra

6

New contributor

Avishek Mitra is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

edited Mar 28 at 10:06

Sujit Bhattacharyya

1,632519

edited Mar 28 at 10:06

Sujit Bhattacharyya

1,632519

asked Mar 28 at 9:55

Avishek MitraAvishek Mitra

6

New contributor

Avishek Mitra is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked Mar 28 at 9:55

Avishek MitraAvishek Mitra

6

2 Answers
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2

$begingroup$

Here is a counterexample. Take the right-angled triangle with sides $(a,b,c)=(3,4,5)$. Then
$$
left(fraca+bcright)^3+left(fracb+caright)^3+left(fracc+abright)^3=frac343125+27+8>24.
$$

share|cite|improve this answer

answered Mar 28 at 10:00

Dietrich BurdeDietrich Burde

81.5k648106

$endgroup$

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  In fact, by taking an isosceles triangle , and making the base angles as close to ninety as desired, one of the above ratios on the left can be made as large as desired!
  $endgroup$
  – астон вілла олоф мэллбэрг
  Mar 28 at 10:03
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Since the value for an equilateral triangle is $24,$ I wonder if the problem should say $"geq24"$ instead of $"<24"$
  $endgroup$
  – saulspatz
  Mar 28 at 10:54
  
  

  
  
  
  

add a comment | 

2

$begingroup$

Actually, the reverse inequality is good : indeed, use the convexity of $x mapsto x^3$. You get that
$$ left( frac13 times fraca+bc + frac13 times fracb+ca + frac13 times fraca+cbright)^3 leq frac13 left(fraca+bcright)^3 + frac13 left(fracb+caright)^3 + frac13 left(fraca+cbright)^3$$

i.e.
$$frac127left( fraca+bc + fracb+ca + fracc+abright)^3 leq frac13 left(left(fraca+bcright)^3 + left(fracb+caright)^3 + left(fraca+cbright)^3 right)$$

i.e.

$$frac19 left( fraca^2b + ab^2 + cb^2 + bc^2 +ac^2+ca^2abcright)^3 leq left(fraca+bcright)^3 + left(fracb+caright)^3 + left(fraca+cbright)^3$$

i.e.
$$frac19 left( frac(a+b)(b+c)(c+a)-2abcabcright)^3 leq left(fraca+bcright)^3 + left(fracb+caright)^3 + left(fraca+cbright)^3$$

Now you use the famous inequality, standing for all non negative real numbers :
$$ (a+b)(b+c)(c+a) geq 8abc$$

You get finally
$$frac19 left( frac8abc-2abcabcright)^3 leq left(fraca+bcright)^3 + left(fracb+caright)^3 + left(fraca+cbright)^3$$

so
$$ left(fraca+bcright)^3 + left(fracb+caright)^3 + left(fraca+cbright)^3 geq frac6^39 = 24.$$

One can remark that this inequality stands for all $a$, $b$, $c$ positive, even if they are not the sides of a triangle.

share|cite|improve this answer

answered Mar 28 at 12:01

TheSilverDoeTheSilverDoe

4,997215

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I had proved this myself, by brute force and awkwardness, and was just about to post a question asking for an elegant proof, but now I don't have to.
  $endgroup$
  – saulspatz
  Mar 28 at 12:09
  

  
  
  
  

add a comment | 

2

$begingroup$

Here is a counterexample. Take the right-angled triangle with sides $(a,b,c)=(3,4,5)$. Then
$$
left(fraca+bcright)^3+left(fracb+caright)^3+left(fracc+abright)^3=frac343125+27+8>24.
$$

share|cite|improve this answer

answered Mar 28 at 10:00

Dietrich BurdeDietrich Burde

81.5k648106

$endgroup$

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  In fact, by taking an isosceles triangle , and making the base angles as close to ninety as desired, one of the above ratios on the left can be made as large as desired!
  $endgroup$
  – астон вілла олоф мэллбэрг
  Mar 28 at 10:03
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Since the value for an equilateral triangle is $24,$ I wonder if the problem should say $"geq24"$ instead of $"<24"$
  $endgroup$
  – saulspatz
  Mar 28 at 10:54
  
  

  
  
  
  

add a comment | 

2

$begingroup$

Here is a counterexample. Take the right-angled triangle with sides $(a,b,c)=(3,4,5)$. Then
$$
left(fraca+bcright)^3+left(fracb+caright)^3+left(fracc+abright)^3=frac343125+27+8>24.
$$

share|cite|improve this answer

answered Mar 28 at 10:00

Dietrich BurdeDietrich Burde

81.5k648106

$endgroup$

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  In fact, by taking an isosceles triangle , and making the base angles as close to ninety as desired, one of the above ratios on the left can be made as large as desired!
  $endgroup$
  – астон вілла олоф мэллбэрг
  Mar 28 at 10:03
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Since the value for an equilateral triangle is $24,$ I wonder if the problem should say $"geq24"$ instead of $"<24"$
  $endgroup$
  – saulspatz
  Mar 28 at 10:54
  
  

  
  
  
  

add a comment | 

$begingroup$

Here is a counterexample. Take the right-angled triangle with sides $(a,b,c)=(3,4,5)$. Then
$$
left(fraca+bcright)^3+left(fracb+caright)^3+left(fracc+abright)^3=frac343125+27+8>24.
$$

share|cite|improve this answer

answered Mar 28 at 10:00

Dietrich BurdeDietrich Burde

81.5k648106

$endgroup$

share|cite|improve this answer

answered Mar 28 at 10:00

Dietrich BurdeDietrich Burde

81.5k648106

answered Mar 28 at 10:00

Dietrich BurdeDietrich Burde

81.5k648106

answered Mar 28 at 10:00

Dietrich BurdeDietrich Burde

81.5k648106

2

$begingroup$

Actually, the reverse inequality is good : indeed, use the convexity of $x mapsto x^3$. You get that
$$ left( frac13 times fraca+bc + frac13 times fracb+ca + frac13 times fraca+cbright)^3 leq frac13 left(fraca+bcright)^3 + frac13 left(fracb+caright)^3 + frac13 left(fraca+cbright)^3$$

i.e.
$$frac127left( fraca+bc + fracb+ca + fracc+abright)^3 leq frac13 left(left(fraca+bcright)^3 + left(fracb+caright)^3 + left(fraca+cbright)^3 right)$$

i.e.

$$frac19 left( fraca^2b + ab^2 + cb^2 + bc^2 +ac^2+ca^2abcright)^3 leq left(fraca+bcright)^3 + left(fracb+caright)^3 + left(fraca+cbright)^3$$

i.e.
$$frac19 left( frac(a+b)(b+c)(c+a)-2abcabcright)^3 leq left(fraca+bcright)^3 + left(fracb+caright)^3 + left(fraca+cbright)^3$$

Now you use the famous inequality, standing for all non negative real numbers :
$$ (a+b)(b+c)(c+a) geq 8abc$$

You get finally
$$frac19 left( frac8abc-2abcabcright)^3 leq left(fraca+bcright)^3 + left(fracb+caright)^3 + left(fraca+cbright)^3$$

so
$$ left(fraca+bcright)^3 + left(fracb+caright)^3 + left(fraca+cbright)^3 geq frac6^39 = 24.$$

One can remark that this inequality stands for all $a$, $b$, $c$ positive, even if they are not the sides of a triangle.

share|cite|improve this answer

answered Mar 28 at 12:01

TheSilverDoeTheSilverDoe

4,997215

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I had proved this myself, by brute force and awkwardness, and was just about to post a question asking for an elegant proof, but now I don't have to.
  $endgroup$
  – saulspatz
  Mar 28 at 12:09
  

  
  
  
  

add a comment | 

2

$begingroup$

Actually, the reverse inequality is good : indeed, use the convexity of $x mapsto x^3$. You get that
$$ left( frac13 times fraca+bc + frac13 times fracb+ca + frac13 times fraca+cbright)^3 leq frac13 left(fraca+bcright)^3 + frac13 left(fracb+caright)^3 + frac13 left(fraca+cbright)^3$$

i.e.
$$frac127left( fraca+bc + fracb+ca + fracc+abright)^3 leq frac13 left(left(fraca+bcright)^3 + left(fracb+caright)^3 + left(fraca+cbright)^3 right)$$

i.e.

$$frac19 left( fraca^2b + ab^2 + cb^2 + bc^2 +ac^2+ca^2abcright)^3 leq left(fraca+bcright)^3 + left(fracb+caright)^3 + left(fraca+cbright)^3$$

i.e.
$$frac19 left( frac(a+b)(b+c)(c+a)-2abcabcright)^3 leq left(fraca+bcright)^3 + left(fracb+caright)^3 + left(fraca+cbright)^3$$

Now you use the famous inequality, standing for all non negative real numbers :
$$ (a+b)(b+c)(c+a) geq 8abc$$

You get finally
$$frac19 left( frac8abc-2abcabcright)^3 leq left(fraca+bcright)^3 + left(fracb+caright)^3 + left(fraca+cbright)^3$$

so
$$ left(fraca+bcright)^3 + left(fracb+caright)^3 + left(fraca+cbright)^3 geq frac6^39 = 24.$$

One can remark that this inequality stands for all $a$, $b$, $c$ positive, even if they are not the sides of a triangle.

share|cite|improve this answer

answered Mar 28 at 12:01

TheSilverDoeTheSilverDoe

4,997215

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I had proved this myself, by brute force and awkwardness, and was just about to post a question asking for an elegant proof, but now I don't have to.
  $endgroup$
  – saulspatz
  Mar 28 at 12:09
  

  
  
  
  

add a comment | 

$begingroup$

Actually, the reverse inequality is good : indeed, use the convexity of $x mapsto x^3$. You get that
$$ left( frac13 times fraca+bc + frac13 times fracb+ca + frac13 times fraca+cbright)^3 leq frac13 left(fraca+bcright)^3 + frac13 left(fracb+caright)^3 + frac13 left(fraca+cbright)^3$$

i.e.
$$frac127left( fraca+bc + fracb+ca + fracc+abright)^3 leq frac13 left(left(fraca+bcright)^3 + left(fracb+caright)^3 + left(fraca+cbright)^3 right)$$

i.e.

$$frac19 left( fraca^2b + ab^2 + cb^2 + bc^2 +ac^2+ca^2abcright)^3 leq left(fraca+bcright)^3 + left(fracb+caright)^3 + left(fraca+cbright)^3$$

i.e.
$$frac19 left( frac(a+b)(b+c)(c+a)-2abcabcright)^3 leq left(fraca+bcright)^3 + left(fracb+caright)^3 + left(fraca+cbright)^3$$

Now you use the famous inequality, standing for all non negative real numbers :
$$ (a+b)(b+c)(c+a) geq 8abc$$

You get finally
$$frac19 left( frac8abc-2abcabcright)^3 leq left(fraca+bcright)^3 + left(fracb+caright)^3 + left(fraca+cbright)^3$$

so
$$ left(fraca+bcright)^3 + left(fracb+caright)^3 + left(fraca+cbright)^3 geq frac6^39 = 24.$$

One can remark that this inequality stands for all $a$, $b$, $c$ positive, even if they are not the sides of a triangle.

share|cite|improve this answer

answered Mar 28 at 12:01

TheSilverDoeTheSilverDoe

4,997215

$endgroup$

share|cite|improve this answer

answered Mar 28 at 12:01

TheSilverDoeTheSilverDoe

4,997215

answered Mar 28 at 12:01

TheSilverDoeTheSilverDoe

4,997215

answered Mar 28 at 12:01

TheSilverDoeTheSilverDoe

4,997215