Inequality in 3 variables with a constraint condition

2

$begingroup$

To prove : a(b+c)/bc + b(c+a)/ca + c(a+b)/ab > 2/(ab+bc+ca) where a+b+c=1 and a,b,c are positive real numbers

Here's my way : Add 1 to each summand in the LHS and subtract 3 (=1+1+1) and after some algebraic manipulation I got (ab+bc+ac)/abc > 2/(ab+bc+ac) +3 which again after some algebra reduces to a²b²+b²c²+a²c² +2abc(a+b+c) > 2+3abc(a+b+c)=> a²b²+b²c²+a²c² > 2+abc . Now, this obviously true for positive a,b,c but I can't prove it rigorously this.

inequality contest-math cauchy-schwarz-inequality

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edited Jan 9 at 22:13

Michael Rozenberg

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asked Jan 9 at 16:31

Epsilon zeroEpsilon zero

35118

$endgroup$

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2

$begingroup$

To prove : a(b+c)/bc + b(c+a)/ca + c(a+b)/ab > 2/(ab+bc+ca) where a+b+c=1 and a,b,c are positive real numbers

Here's my way : Add 1 to each summand in the LHS and subtract 3 (=1+1+1) and after some algebraic manipulation I got (ab+bc+ac)/abc > 2/(ab+bc+ac) +3 which again after some algebra reduces to a²b²+b²c²+a²c² +2abc(a+b+c) > 2+3abc(a+b+c)=> a²b²+b²c²+a²c² > 2+abc . Now, this obviously true for positive a,b,c but I can't prove it rigorously this.

inequality contest-math cauchy-schwarz-inequality

share|cite|improve this question

edited Jan 9 at 22:13

Michael Rozenberg

101k1591193

asked Jan 9 at 16:31

Epsilon zeroEpsilon zero

35118

$endgroup$

add a comment | 

2

2

2

$begingroup$

To prove : a(b+c)/bc + b(c+a)/ca + c(a+b)/ab > 2/(ab+bc+ca) where a+b+c=1 and a,b,c are positive real numbers

Here's my way : Add 1 to each summand in the LHS and subtract 3 (=1+1+1) and after some algebraic manipulation I got (ab+bc+ac)/abc > 2/(ab+bc+ac) +3 which again after some algebra reduces to a²b²+b²c²+a²c² +2abc(a+b+c) > 2+3abc(a+b+c)=> a²b²+b²c²+a²c² > 2+abc . Now, this obviously true for positive a,b,c but I can't prove it rigorously this.

inequality contest-math cauchy-schwarz-inequality

share|cite|improve this question

edited Jan 9 at 22:13

Michael Rozenberg

101k1591193

asked Jan 9 at 16:31

Epsilon zeroEpsilon zero

35118

$endgroup$

To prove : a(b+c)/bc + b(c+a)/ca + c(a+b)/ab > 2/(ab+bc+ca) where a+b+c=1 and a,b,c are positive real numbers

Here's my way : Add 1 to each summand in the LHS and subtract 3 (=1+1+1) and after some algebraic manipulation I got (ab+bc+ac)/abc > 2/(ab+bc+ac) +3 which again after some algebra reduces to a²b²+b²c²+a²c² +2abc(a+b+c) > 2+3abc(a+b+c)=> a²b²+b²c²+a²c² > 2+abc . Now, this obviously true for positive a,b,c but I can't prove it rigorously this.

inequality contest-math cauchy-schwarz-inequality

inequality contest-math cauchy-schwarz-inequality

share|cite|improve this question

edited Jan 9 at 22:13

Michael Rozenberg

101k1591193

asked Jan 9 at 16:31

Epsilon zeroEpsilon zero

35118

share|cite|improve this question

edited Jan 9 at 22:13

Michael Rozenberg

101k1591193

asked Jan 9 at 16:31

Epsilon zeroEpsilon zero

35118

share|cite|improve this question

share|cite|improve this question

edited Jan 9 at 22:13

Michael Rozenberg

101k1591193

edited Jan 9 at 22:13

Michael Rozenberg

101k1591193

edited Jan 9 at 22:13

Michael Rozenberg

101k1591193

101k1591193

asked Jan 9 at 16:31

Epsilon zeroEpsilon zero

35118

asked Jan 9 at 16:31

Epsilon zeroEpsilon zero

35118

asked Jan 9 at 16:31

Epsilon zeroEpsilon zero

35118

35118

add a comment | 

add a comment | 

2 Answers
2

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oldest

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2

$begingroup$

If we arrange
$$
frac{ab+bc+ac}{abc} ge frac{2}{ab+bc+ac} +3
$$ then we have
$$
(ab+bc+ac)^2 =sum_{text{cyc}}a^2b^2 +2abcge abc(2+3(ab+bc+ac))=2abc +3sum_{text{cyc}}a^2b^2c.
$$ We need to show
$$
sum_{text{cyc}}a^2b^2ge 3sum_{text{cyc}}a^2b^2c.
$$ This is equivalent to
$$
sum_{text{cyc}}a^2b^2(a+b+c)ge 3sum_{text{cyc}}a^2b^2c.
$$ By rearrangement, we have
$$
sum_{text{cyc}}a^3b^2ge sum_{text{cyc}}a^2b^2c,
$$and
$$
sum_{text{cyc}}a^2b^3ge sum_{text{cyc}}a^2b^2c.
$$ This proves the inequality.

Another approach: We need to show
$$
sum_{text{cyc}}frac{a}{b}+sum_{text{cyc}}frac{a}{c}ge frac{2}{ab+bc+ca}.
$$ By C-S, we have
$$
sum_{text{cyc}}frac{a}{b}sum_{text{cyc}}abge left(sum_{text{cyc}}aright)^2 =1,
$$ and
$$
sum_{text{cyc}}frac{a}{c}sum_{text{cyc}}acge left(sum_{text{cyc}}aright)^2 =1.
$$

share|cite|improve this answer

edited Jan 9 at 17:21

answered Jan 9 at 17:14

SongSong

11.4k628

$endgroup$

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0

$begingroup$

Because by C-S
$$(ab+ac+bc)sum_{cyc}frac{a(b+c)}{bc}=sum_{cyc}acsum_{cyc}frac{a}{c}+sum_{cyc}absum_{cyc}frac{a}{b}geq$$
$$geq(a+b+c)^2+(a+b+c)^2=2.$$

share|cite|improve this answer

answered Jan 9 at 22:13

Michael RozenbergMichael Rozenberg

101k1591193

$endgroup$

add a comment | 

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

2

$begingroup$

If we arrange
$$
frac{ab+bc+ac}{abc} ge frac{2}{ab+bc+ac} +3
$$ then we have
$$
(ab+bc+ac)^2 =sum_{text{cyc}}a^2b^2 +2abcge abc(2+3(ab+bc+ac))=2abc +3sum_{text{cyc}}a^2b^2c.
$$ We need to show
$$
sum_{text{cyc}}a^2b^2ge 3sum_{text{cyc}}a^2b^2c.
$$ This is equivalent to
$$
sum_{text{cyc}}a^2b^2(a+b+c)ge 3sum_{text{cyc}}a^2b^2c.
$$ By rearrangement, we have
$$
sum_{text{cyc}}a^3b^2ge sum_{text{cyc}}a^2b^2c,
$$and
$$
sum_{text{cyc}}a^2b^3ge sum_{text{cyc}}a^2b^2c.
$$ This proves the inequality.

Another approach: We need to show
$$
sum_{text{cyc}}frac{a}{b}+sum_{text{cyc}}frac{a}{c}ge frac{2}{ab+bc+ca}.
$$ By C-S, we have
$$
sum_{text{cyc}}frac{a}{b}sum_{text{cyc}}abge left(sum_{text{cyc}}aright)^2 =1,
$$ and
$$
sum_{text{cyc}}frac{a}{c}sum_{text{cyc}}acge left(sum_{text{cyc}}aright)^2 =1.
$$

share|cite|improve this answer

edited Jan 9 at 17:21

answered Jan 9 at 17:14

SongSong

11.4k628

$endgroup$

add a comment | 

2

$begingroup$

If we arrange
$$
frac{ab+bc+ac}{abc} ge frac{2}{ab+bc+ac} +3
$$ then we have
$$
(ab+bc+ac)^2 =sum_{text{cyc}}a^2b^2 +2abcge abc(2+3(ab+bc+ac))=2abc +3sum_{text{cyc}}a^2b^2c.
$$ We need to show
$$
sum_{text{cyc}}a^2b^2ge 3sum_{text{cyc}}a^2b^2c.
$$ This is equivalent to
$$
sum_{text{cyc}}a^2b^2(a+b+c)ge 3sum_{text{cyc}}a^2b^2c.
$$ By rearrangement, we have
$$
sum_{text{cyc}}a^3b^2ge sum_{text{cyc}}a^2b^2c,
$$and
$$
sum_{text{cyc}}a^2b^3ge sum_{text{cyc}}a^2b^2c.
$$ This proves the inequality.

Another approach: We need to show
$$
sum_{text{cyc}}frac{a}{b}+sum_{text{cyc}}frac{a}{c}ge frac{2}{ab+bc+ca}.
$$ By C-S, we have
$$
sum_{text{cyc}}frac{a}{b}sum_{text{cyc}}abge left(sum_{text{cyc}}aright)^2 =1,
$$ and
$$
sum_{text{cyc}}frac{a}{c}sum_{text{cyc}}acge left(sum_{text{cyc}}aright)^2 =1.
$$

share|cite|improve this answer

edited Jan 9 at 17:21

answered Jan 9 at 17:14

SongSong

11.4k628

$endgroup$

add a comment | 

2

2

2

$begingroup$

If we arrange
$$
frac{ab+bc+ac}{abc} ge frac{2}{ab+bc+ac} +3
$$ then we have
$$
(ab+bc+ac)^2 =sum_{text{cyc}}a^2b^2 +2abcge abc(2+3(ab+bc+ac))=2abc +3sum_{text{cyc}}a^2b^2c.
$$ We need to show
$$
sum_{text{cyc}}a^2b^2ge 3sum_{text{cyc}}a^2b^2c.
$$ This is equivalent to
$$
sum_{text{cyc}}a^2b^2(a+b+c)ge 3sum_{text{cyc}}a^2b^2c.
$$ By rearrangement, we have
$$
sum_{text{cyc}}a^3b^2ge sum_{text{cyc}}a^2b^2c,
$$and
$$
sum_{text{cyc}}a^2b^3ge sum_{text{cyc}}a^2b^2c.
$$ This proves the inequality.

Another approach: We need to show
$$
sum_{text{cyc}}frac{a}{b}+sum_{text{cyc}}frac{a}{c}ge frac{2}{ab+bc+ca}.
$$ By C-S, we have
$$
sum_{text{cyc}}frac{a}{b}sum_{text{cyc}}abge left(sum_{text{cyc}}aright)^2 =1,
$$ and
$$
sum_{text{cyc}}frac{a}{c}sum_{text{cyc}}acge left(sum_{text{cyc}}aright)^2 =1.
$$

share|cite|improve this answer

edited Jan 9 at 17:21

answered Jan 9 at 17:14

SongSong

11.4k628

$endgroup$

If we arrange
$$
frac{ab+bc+ac}{abc} ge frac{2}{ab+bc+ac} +3
$$ then we have
$$
(ab+bc+ac)^2 =sum_{text{cyc}}a^2b^2 +2abcge abc(2+3(ab+bc+ac))=2abc +3sum_{text{cyc}}a^2b^2c.
$$ We need to show
$$
sum_{text{cyc}}a^2b^2ge 3sum_{text{cyc}}a^2b^2c.
$$ This is equivalent to
$$
sum_{text{cyc}}a^2b^2(a+b+c)ge 3sum_{text{cyc}}a^2b^2c.
$$ By rearrangement, we have
$$
sum_{text{cyc}}a^3b^2ge sum_{text{cyc}}a^2b^2c,
$$and
$$
sum_{text{cyc}}a^2b^3ge sum_{text{cyc}}a^2b^2c.
$$ This proves the inequality.

Another approach: We need to show
$$
sum_{text{cyc}}frac{a}{b}+sum_{text{cyc}}frac{a}{c}ge frac{2}{ab+bc+ca}.
$$ By C-S, we have
$$
sum_{text{cyc}}frac{a}{b}sum_{text{cyc}}abge left(sum_{text{cyc}}aright)^2 =1,
$$ and
$$
sum_{text{cyc}}frac{a}{c}sum_{text{cyc}}acge left(sum_{text{cyc}}aright)^2 =1.
$$

share|cite|improve this answer

edited Jan 9 at 17:21

answered Jan 9 at 17:14

SongSong

11.4k628

share|cite|improve this answer

share|cite|improve this answer

edited Jan 9 at 17:21

edited Jan 9 at 17:21

edited Jan 9 at 17:21

answered Jan 9 at 17:14

SongSong

11.4k628

answered Jan 9 at 17:14

SongSong

11.4k628

answered Jan 9 at 17:14

SongSong

11.4k628

11.4k628

add a comment | 

add a comment | 

0

$begingroup$

Because by C-S
$$(ab+ac+bc)sum_{cyc}frac{a(b+c)}{bc}=sum_{cyc}acsum_{cyc}frac{a}{c}+sum_{cyc}absum_{cyc}frac{a}{b}geq$$
$$geq(a+b+c)^2+(a+b+c)^2=2.$$

share|cite|improve this answer

answered Jan 9 at 22:13

Michael RozenbergMichael Rozenberg

101k1591193

$endgroup$

add a comment | 

0

$begingroup$

Because by C-S
$$(ab+ac+bc)sum_{cyc}frac{a(b+c)}{bc}=sum_{cyc}acsum_{cyc}frac{a}{c}+sum_{cyc}absum_{cyc}frac{a}{b}geq$$
$$geq(a+b+c)^2+(a+b+c)^2=2.$$

share|cite|improve this answer

answered Jan 9 at 22:13

Michael RozenbergMichael Rozenberg

101k1591193

$endgroup$

add a comment | 

0

0

0

$begingroup$

Because by C-S
$$(ab+ac+bc)sum_{cyc}frac{a(b+c)}{bc}=sum_{cyc}acsum_{cyc}frac{a}{c}+sum_{cyc}absum_{cyc}frac{a}{b}geq$$
$$geq(a+b+c)^2+(a+b+c)^2=2.$$

share|cite|improve this answer

answered Jan 9 at 22:13

Michael RozenbergMichael Rozenberg

101k1591193

$endgroup$

Because by C-S
$$(ab+ac+bc)sum_{cyc}frac{a(b+c)}{bc}=sum_{cyc}acsum_{cyc}frac{a}{c}+sum_{cyc}absum_{cyc}frac{a}{b}geq$$
$$geq(a+b+c)^2+(a+b+c)^2=2.$$

share|cite|improve this answer

answered Jan 9 at 22:13

Michael RozenbergMichael Rozenberg

101k1591193

share|cite|improve this answer

share|cite|improve this answer

answered Jan 9 at 22:13

Michael RozenbergMichael Rozenberg

101k1591193

answered Jan 9 at 22:13

Michael RozenbergMichael Rozenberg

101k1591193

answered Jan 9 at 22:13

Michael RozenbergMichael Rozenberg

101k1591193

101k1591193

add a comment | 

add a comment | 

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Use MathJax to format equations. MathJax reference.

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Use MathJax to format equations. MathJax reference.

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