If $a,b,c$ are positive real numbers and $(1+a+b+c)left(1+frac1a+frac1b+frac1cright)=16$, then $a+b+c=3$ [duplicate]

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











up vote
0
down vote

favorite
1













This question already has an answer here:



  • Condition on a,b and c satisfying an equation(TIFR GS 2017)

    3 answers




Let $a,b,c$ be positive real number such that
$$(1+a+b+c)left(1+frac1a+frac1b+frac1cright)=16$$
then $a+b+c=3$




I think this is trivial but I don't know how to prove this fact.



Any Help will be appreciated










share|cite|improve this question















marked as duplicate by Chinnapparaj R, Lord Shark the Unknown, user21820, Henry, José Carlos Santos real-analysis
Users with the  real-analysis badge can single-handedly close real-analysis questions as duplicates and reopen them as needed.

StackExchange.ready(function()
if (StackExchange.options.isMobile) return;

$('.dupe-hammer-message-hover:not(.hover-bound)').each(function()
var $hover = $(this).addClass('hover-bound'),
$msg = $hover.siblings('.dupe-hammer-message');

$hover.hover(
function()
$hover.showInfoMessage('',
messageElement: $msg.clone().show(),
transient: false,
position: my: 'bottom left', at: 'top center', offsetTop: -7 ,
dismissable: false,
relativeToBody: true
);
,
function()
StackExchange.helpers.removeMessages();

);
);
);
Nov 29 at 9:53


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.














  • it must be wrong, let b=1, c=1, it is clear that the solution for a is not 1
    – MoonKnight
    Nov 29 at 5:59






  • 1




    Do you mean $(1+a+b+c)(1+1/a+1/b+1/c)=16$?
    – Lord Shark the Unknown
    Nov 29 at 6:00










  • @MathLover: I edit your question. I remove the plus sign in the middle. Because if it were then this result is wrong.If you feel I'm missing something, kindly go ahead and edit again!
    – Chinnapparaj R
    Nov 29 at 6:05










  • You have edited right ...Thanks a lot
    – MathLover
    Nov 29 at 6:07










  • Your question is already appeared on this site. Hold on ....I mention that link!
    – Chinnapparaj R
    Nov 29 at 6:19














up vote
0
down vote

favorite
1













This question already has an answer here:



  • Condition on a,b and c satisfying an equation(TIFR GS 2017)

    3 answers




Let $a,b,c$ be positive real number such that
$$(1+a+b+c)left(1+frac1a+frac1b+frac1cright)=16$$
then $a+b+c=3$




I think this is trivial but I don't know how to prove this fact.



Any Help will be appreciated










share|cite|improve this question















marked as duplicate by Chinnapparaj R, Lord Shark the Unknown, user21820, Henry, José Carlos Santos real-analysis
Users with the  real-analysis badge can single-handedly close real-analysis questions as duplicates and reopen them as needed.

StackExchange.ready(function()
if (StackExchange.options.isMobile) return;

$('.dupe-hammer-message-hover:not(.hover-bound)').each(function()
var $hover = $(this).addClass('hover-bound'),
$msg = $hover.siblings('.dupe-hammer-message');

$hover.hover(
function()
$hover.showInfoMessage('',
messageElement: $msg.clone().show(),
transient: false,
position: my: 'bottom left', at: 'top center', offsetTop: -7 ,
dismissable: false,
relativeToBody: true
);
,
function()
StackExchange.helpers.removeMessages();

);
);
);
Nov 29 at 9:53


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.














  • it must be wrong, let b=1, c=1, it is clear that the solution for a is not 1
    – MoonKnight
    Nov 29 at 5:59






  • 1




    Do you mean $(1+a+b+c)(1+1/a+1/b+1/c)=16$?
    – Lord Shark the Unknown
    Nov 29 at 6:00










  • @MathLover: I edit your question. I remove the plus sign in the middle. Because if it were then this result is wrong.If you feel I'm missing something, kindly go ahead and edit again!
    – Chinnapparaj R
    Nov 29 at 6:05










  • You have edited right ...Thanks a lot
    – MathLover
    Nov 29 at 6:07










  • Your question is already appeared on this site. Hold on ....I mention that link!
    – Chinnapparaj R
    Nov 29 at 6:19












up vote
0
down vote

favorite
1









up vote
0
down vote

favorite
1






1






This question already has an answer here:



  • Condition on a,b and c satisfying an equation(TIFR GS 2017)

    3 answers




Let $a,b,c$ be positive real number such that
$$(1+a+b+c)left(1+frac1a+frac1b+frac1cright)=16$$
then $a+b+c=3$




I think this is trivial but I don't know how to prove this fact.



Any Help will be appreciated










share|cite|improve this question
















This question already has an answer here:



  • Condition on a,b and c satisfying an equation(TIFR GS 2017)

    3 answers




Let $a,b,c$ be positive real number such that
$$(1+a+b+c)left(1+frac1a+frac1b+frac1cright)=16$$
then $a+b+c=3$




I think this is trivial but I don't know how to prove this fact.



Any Help will be appreciated





This question already has an answer here:



  • Condition on a,b and c satisfying an equation(TIFR GS 2017)

    3 answers







real-analysis analysis






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 29 at 8:45









Asaf Karagila

300k32422751




300k32422751










asked Nov 29 at 5:56









MathLover

44010




44010




marked as duplicate by Chinnapparaj R, Lord Shark the Unknown, user21820, Henry, José Carlos Santos real-analysis
Users with the  real-analysis badge can single-handedly close real-analysis questions as duplicates and reopen them as needed.

StackExchange.ready(function()
if (StackExchange.options.isMobile) return;

$('.dupe-hammer-message-hover:not(.hover-bound)').each(function()
var $hover = $(this).addClass('hover-bound'),
$msg = $hover.siblings('.dupe-hammer-message');

$hover.hover(
function()
$hover.showInfoMessage('',
messageElement: $msg.clone().show(),
transient: false,
position: my: 'bottom left', at: 'top center', offsetTop: -7 ,
dismissable: false,
relativeToBody: true
);
,
function()
StackExchange.helpers.removeMessages();

);
);
);
Nov 29 at 9:53


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.






marked as duplicate by Chinnapparaj R, Lord Shark the Unknown, user21820, Henry, José Carlos Santos real-analysis
Users with the  real-analysis badge can single-handedly close real-analysis questions as duplicates and reopen them as needed.

StackExchange.ready(function()
if (StackExchange.options.isMobile) return;

$('.dupe-hammer-message-hover:not(.hover-bound)').each(function()
var $hover = $(this).addClass('hover-bound'),
$msg = $hover.siblings('.dupe-hammer-message');

$hover.hover(
function()
$hover.showInfoMessage('',
messageElement: $msg.clone().show(),
transient: false,
position: my: 'bottom left', at: 'top center', offsetTop: -7 ,
dismissable: false,
relativeToBody: true
);
,
function()
StackExchange.helpers.removeMessages();

);
);
);
Nov 29 at 9:53


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.













  • it must be wrong, let b=1, c=1, it is clear that the solution for a is not 1
    – MoonKnight
    Nov 29 at 5:59






  • 1




    Do you mean $(1+a+b+c)(1+1/a+1/b+1/c)=16$?
    – Lord Shark the Unknown
    Nov 29 at 6:00










  • @MathLover: I edit your question. I remove the plus sign in the middle. Because if it were then this result is wrong.If you feel I'm missing something, kindly go ahead and edit again!
    – Chinnapparaj R
    Nov 29 at 6:05










  • You have edited right ...Thanks a lot
    – MathLover
    Nov 29 at 6:07










  • Your question is already appeared on this site. Hold on ....I mention that link!
    – Chinnapparaj R
    Nov 29 at 6:19
















  • it must be wrong, let b=1, c=1, it is clear that the solution for a is not 1
    – MoonKnight
    Nov 29 at 5:59






  • 1




    Do you mean $(1+a+b+c)(1+1/a+1/b+1/c)=16$?
    – Lord Shark the Unknown
    Nov 29 at 6:00










  • @MathLover: I edit your question. I remove the plus sign in the middle. Because if it were then this result is wrong.If you feel I'm missing something, kindly go ahead and edit again!
    – Chinnapparaj R
    Nov 29 at 6:05










  • You have edited right ...Thanks a lot
    – MathLover
    Nov 29 at 6:07










  • Your question is already appeared on this site. Hold on ....I mention that link!
    – Chinnapparaj R
    Nov 29 at 6:19















it must be wrong, let b=1, c=1, it is clear that the solution for a is not 1
– MoonKnight
Nov 29 at 5:59




it must be wrong, let b=1, c=1, it is clear that the solution for a is not 1
– MoonKnight
Nov 29 at 5:59




1




1




Do you mean $(1+a+b+c)(1+1/a+1/b+1/c)=16$?
– Lord Shark the Unknown
Nov 29 at 6:00




Do you mean $(1+a+b+c)(1+1/a+1/b+1/c)=16$?
– Lord Shark the Unknown
Nov 29 at 6:00












@MathLover: I edit your question. I remove the plus sign in the middle. Because if it were then this result is wrong.If you feel I'm missing something, kindly go ahead and edit again!
– Chinnapparaj R
Nov 29 at 6:05




@MathLover: I edit your question. I remove the plus sign in the middle. Because if it were then this result is wrong.If you feel I'm missing something, kindly go ahead and edit again!
– Chinnapparaj R
Nov 29 at 6:05












You have edited right ...Thanks a lot
– MathLover
Nov 29 at 6:07




You have edited right ...Thanks a lot
– MathLover
Nov 29 at 6:07












Your question is already appeared on this site. Hold on ....I mention that link!
– Chinnapparaj R
Nov 29 at 6:19




Your question is already appeared on this site. Hold on ....I mention that link!
– Chinnapparaj R
Nov 29 at 6:19










3 Answers
3






active

oldest

votes

















up vote
3
down vote













Use the AM-HM inequality of $1, a, b, c$.



So $$frac1 + a + b + c4 geq frac41 + frac1a + frac1b + frac1c$$



This gives $$(1 + a + b + c)bigg(1 + frac1a + frac1b + frac1cbigg) geq 16$$.



The equality becomes an equation only when all four elements are equal.



i.e. $1=a=b=c$. So $a + b + c = 3$.






share|cite|improve this answer





























    up vote
    3
    down vote













    Hint:



    Using AM HM inequality for non-negative numbers,



    $$dfracsum_r=1^na_rngedfrac nsum_r=1^ndfrac1a_r$$






    share|cite|improve this answer





























      up vote
      2
      down vote













      Using Cauchy-Schwarz inequality you can even show that $a=b=c=1$:
      $$(1+a+b+c)left(1+frac1a+frac1b+frac1cright) = left(1^2 + sum_cycleft(sqrtaright)^2 right)left(1^2 + sum_cycleft(frac1sqrtaright)^2 right)$$ $$stackrelC.-S.geq(1+1+1+1)^2 = 16$$
      Equality holds for $(1: sqrta : sqrtb : sqrtc)^T = lambda cdot left( 1: frac1sqrta : : frac1sqrtb : : frac1sqrtcright)^T Rightarrow a=b=c = 1$






      share|cite|improve this answer





























        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes








        up vote
        3
        down vote













        Use the AM-HM inequality of $1, a, b, c$.



        So $$frac1 + a + b + c4 geq frac41 + frac1a + frac1b + frac1c$$



        This gives $$(1 + a + b + c)bigg(1 + frac1a + frac1b + frac1cbigg) geq 16$$.



        The equality becomes an equation only when all four elements are equal.



        i.e. $1=a=b=c$. So $a + b + c = 3$.






        share|cite|improve this answer


























          up vote
          3
          down vote













          Use the AM-HM inequality of $1, a, b, c$.



          So $$frac1 + a + b + c4 geq frac41 + frac1a + frac1b + frac1c$$



          This gives $$(1 + a + b + c)bigg(1 + frac1a + frac1b + frac1cbigg) geq 16$$.



          The equality becomes an equation only when all four elements are equal.



          i.e. $1=a=b=c$. So $a + b + c = 3$.






          share|cite|improve this answer
























            up vote
            3
            down vote










            up vote
            3
            down vote









            Use the AM-HM inequality of $1, a, b, c$.



            So $$frac1 + a + b + c4 geq frac41 + frac1a + frac1b + frac1c$$



            This gives $$(1 + a + b + c)bigg(1 + frac1a + frac1b + frac1cbigg) geq 16$$.



            The equality becomes an equation only when all four elements are equal.



            i.e. $1=a=b=c$. So $a + b + c = 3$.






            share|cite|improve this answer














            Use the AM-HM inequality of $1, a, b, c$.



            So $$frac1 + a + b + c4 geq frac41 + frac1a + frac1b + frac1c$$



            This gives $$(1 + a + b + c)bigg(1 + frac1a + frac1b + frac1cbigg) geq 16$$.



            The equality becomes an equation only when all four elements are equal.



            i.e. $1=a=b=c$. So $a + b + c = 3$.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Nov 29 at 6:54









            KM101

            3,416417




            3,416417










            answered Nov 29 at 6:48









            Bhargav Kale

            412




            412




















                up vote
                3
                down vote













                Hint:



                Using AM HM inequality for non-negative numbers,



                $$dfracsum_r=1^na_rngedfrac nsum_r=1^ndfrac1a_r$$






                share|cite|improve this answer


























                  up vote
                  3
                  down vote













                  Hint:



                  Using AM HM inequality for non-negative numbers,



                  $$dfracsum_r=1^na_rngedfrac nsum_r=1^ndfrac1a_r$$






                  share|cite|improve this answer
























                    up vote
                    3
                    down vote










                    up vote
                    3
                    down vote









                    Hint:



                    Using AM HM inequality for non-negative numbers,



                    $$dfracsum_r=1^na_rngedfrac nsum_r=1^ndfrac1a_r$$






                    share|cite|improve this answer














                    Hint:



                    Using AM HM inequality for non-negative numbers,



                    $$dfracsum_r=1^na_rngedfrac nsum_r=1^ndfrac1a_r$$







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Nov 29 at 7:04

























                    answered Nov 29 at 5:59









                    lab bhattacharjee

                    221k15155273




                    221k15155273




















                        up vote
                        2
                        down vote













                        Using Cauchy-Schwarz inequality you can even show that $a=b=c=1$:
                        $$(1+a+b+c)left(1+frac1a+frac1b+frac1cright) = left(1^2 + sum_cycleft(sqrtaright)^2 right)left(1^2 + sum_cycleft(frac1sqrtaright)^2 right)$$ $$stackrelC.-S.geq(1+1+1+1)^2 = 16$$
                        Equality holds for $(1: sqrta : sqrtb : sqrtc)^T = lambda cdot left( 1: frac1sqrta : : frac1sqrtb : : frac1sqrtcright)^T Rightarrow a=b=c = 1$






                        share|cite|improve this answer


























                          up vote
                          2
                          down vote













                          Using Cauchy-Schwarz inequality you can even show that $a=b=c=1$:
                          $$(1+a+b+c)left(1+frac1a+frac1b+frac1cright) = left(1^2 + sum_cycleft(sqrtaright)^2 right)left(1^2 + sum_cycleft(frac1sqrtaright)^2 right)$$ $$stackrelC.-S.geq(1+1+1+1)^2 = 16$$
                          Equality holds for $(1: sqrta : sqrtb : sqrtc)^T = lambda cdot left( 1: frac1sqrta : : frac1sqrtb : : frac1sqrtcright)^T Rightarrow a=b=c = 1$






                          share|cite|improve this answer
























                            up vote
                            2
                            down vote










                            up vote
                            2
                            down vote









                            Using Cauchy-Schwarz inequality you can even show that $a=b=c=1$:
                            $$(1+a+b+c)left(1+frac1a+frac1b+frac1cright) = left(1^2 + sum_cycleft(sqrtaright)^2 right)left(1^2 + sum_cycleft(frac1sqrtaright)^2 right)$$ $$stackrelC.-S.geq(1+1+1+1)^2 = 16$$
                            Equality holds for $(1: sqrta : sqrtb : sqrtc)^T = lambda cdot left( 1: frac1sqrta : : frac1sqrtb : : frac1sqrtcright)^T Rightarrow a=b=c = 1$






                            share|cite|improve this answer














                            Using Cauchy-Schwarz inequality you can even show that $a=b=c=1$:
                            $$(1+a+b+c)left(1+frac1a+frac1b+frac1cright) = left(1^2 + sum_cycleft(sqrtaright)^2 right)left(1^2 + sum_cycleft(frac1sqrtaright)^2 right)$$ $$stackrelC.-S.geq(1+1+1+1)^2 = 16$$
                            Equality holds for $(1: sqrta : sqrtb : sqrtc)^T = lambda cdot left( 1: frac1sqrta : : frac1sqrtb : : frac1sqrtcright)^T Rightarrow a=b=c = 1$







                            share|cite|improve this answer














                            share|cite|improve this answer



                            share|cite|improve this answer








                            edited Nov 29 at 6:53

























                            answered Nov 29 at 6:48









                            trancelocation

                            8,6321520




                            8,6321520












                                Popular posts from this blog

                                How to check contact read email or not when send email to Individual?

                                Displaying single band from multi-band raster using QGIS

                                How many registers does an x86_64 CPU actually have?