finding minimum value of the expression












1















What is the minimum value of the expression



$$(a+1/a)^2+(b+1/b)^2+(c+1/c)^2$$




How do we go ahead solving this problem ? Can we use $AMge HM$ to solve this problem ?The answer is supposed to be 12 which i can see in answer booklet but i have no clue how to solve this even when i know the concept of AM>=HM.






inequality optimization arithmetic







share|cite|improve this question
























  • And $$a,b,c$$ assumed to be positive?
    – Dr. Sonnhard Graubner
    1 hour ago










  • $a = b = c = -1$ or $1$.
    – David G. Stork
    1 hour ago












  • I think it must be additionally $$a+b+c=1$$ and $$a>0,b>0,c>0$$
    – Dr. Sonnhard Graubner
    1 hour ago
















1















What is the minimum value of the expression



$$(a+1/a)^2+(b+1/b)^2+(c+1/c)^2$$




How do we go ahead solving this problem ? Can we use $AMge HM$ to solve this problem ?The answer is supposed to be 12 which i can see in answer booklet but i have no clue how to solve this even when i know the concept of AM>=HM.






inequality optimization arithmetic







share|cite|improve this question
























  • And $$a,b,c$$ assumed to be positive?
    – Dr. Sonnhard Graubner
    1 hour ago










  • $a = b = c = -1$ or $1$.
    – David G. Stork
    1 hour ago












  • I think it must be additionally $$a+b+c=1$$ and $$a>0,b>0,c>0$$
    – Dr. Sonnhard Graubner
    1 hour ago














1












1








1








What is the minimum value of the expression



$$(a+1/a)^2+(b+1/b)^2+(c+1/c)^2$$




How do we go ahead solving this problem ? Can we use $AMge HM$ to solve this problem ?The answer is supposed to be 12 which i can see in answer booklet but i have no clue how to solve this even when i know the concept of AM>=HM.






inequality optimization arithmetic







share|cite|improve this question
















What is the minimum value of the expression



$$(a+1/a)^2+(b+1/b)^2+(c+1/c)^2$$




How do we go ahead solving this problem ? Can we use $AMge HM$ to solve this problem ?The answer is supposed to be 12 which i can see in answer booklet but i have no clue how to solve this even when i know the concept of AM>=HM.






inequality optimization arithmetic




inequality optimization arithmetic






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 1 hour ago









greedoid

37.9k114794




37.9k114794










asked 1 hour ago









iamredpanda

123




123












  • And $$a,b,c$$ assumed to be positive?
    – Dr. Sonnhard Graubner
    1 hour ago










  • $a = b = c = -1$ or $1$.
    – David G. Stork
    1 hour ago












  • I think it must be additionally $$a+b+c=1$$ and $$a>0,b>0,c>0$$
    – Dr. Sonnhard Graubner
    1 hour ago


















  • And $$a,b,c$$ assumed to be positive?
    – Dr. Sonnhard Graubner
    1 hour ago










  • $a = b = c = -1$ or $1$.
    – David G. Stork
    1 hour ago












  • I think it must be additionally $$a+b+c=1$$ and $$a>0,b>0,c>0$$
    – Dr. Sonnhard Graubner
    1 hour ago
















And $$a,b,c$$ assumed to be positive?
– Dr. Sonnhard Graubner
1 hour ago




And $$a,b,c$$ assumed to be positive?
– Dr. Sonnhard Graubner
1 hour ago












$a = b = c = -1$ or $1$.
– David G. Stork
1 hour ago






$a = b = c = -1$ or $1$.
– David G. Stork
1 hour ago














I think it must be additionally $$a+b+c=1$$ and $$a>0,b>0,c>0$$
– Dr. Sonnhard Graubner
1 hour ago




I think it must be additionally $$a+b+c=1$$ and $$a>0,b>0,c>0$$
– Dr. Sonnhard Graubner
1 hour ago










3 Answers
3






active

oldest

votes


















5














Since $$(x+1/x)^2 = underbrace{x^2+1/x^2}_{geq 2}+2geq 2+2=4$$ your expression is at least 12
and equality is achieved if $a=b=c=1$.






share|cite|improve this answer





















  • Oh this turned out to be too simple.I was trying to think in wrong direction .
    – iamredpanda
    1 hour ago










  • How do you show $x^2+1/x^2ge2$? Won't you need the same steps that can also show $x+1/xge2$?
    – Teepeemm
    34 mins ago










  • Exactly the same, just write $a=x^2$.
    – greedoid
    33 mins ago



















4














Opimization is another way to solve this.



$$f(a) = a+frac{1}{a} implies f’(a) = 1-a^{-2}$$



Setting $f’(a) = 0$ to obtain the minimum, you have



$$0 = 1-a^{-2} implies a^{-2} = 1 implies a = pm 1$$



(For clarification, $a = -1$ gives the minimum absolute value for negative values of $a$.)



Thus, the minimum is obtained when $a = pm 1$, so you have $left(a+frac{1}{a}right)^2 = (pm 2)^2 = 4$. The same goes for $b$ and $c$, so the minimum occurs when $a = b = c = pm 1$ and is $4+4+4 = 12$.






share|cite|improve this answer























  • why does $f'(a)=0$ coincide with a minimum?
    – LinAlg
    1 hour ago










  • Because the stationary points occur where $f’(a) = 0$.
    – KM101
    1 hour ago






  • 2




    stationary points are not necessarily minima, and the function may not even be bounded below
    – LinAlg
    44 mins ago










  • I guess my reply wasn't clear enough. I didn't mean a stationary point always implies minima, but that it is the case in this example. For instance, for positive $a$, there is $frac{bpm n}{b}+frac{b}{bpm n} = frac{2left(b^2-bnright)+n^2}{b^2-bn}$, so there clearly is a minimum where $n = 0$, so $a = 1$.
    – KM101
    18 mins ago








  • 1




    @Tyberius I addressed that already. Since $a+frac{1}{a} < 0$ for $a < 0$, then $a = -1$ is a local maximum, and allows for the least absolute value for negative values of $a$.
    – KM101
    17 mins ago





















0














The given expression equals
begin{align*}
a^2 + b^2 + c^2 &+ frac{1}{a^2} + frac{1}{b^2} + frac{1}{c^2} + 6 \
&geq 6 left(a^2 cdot b^2 cdot c^2 cdot frac{1}{a^2}frac{1}{b^2}frac{1}{c^2} right)^{frac{1}{6}} + 6 \
&= 12
end{align*}
where we have used $AM - GM$ inequality.






share|cite|improve this answer





















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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    5














    Since $$(x+1/x)^2 = underbrace{x^2+1/x^2}_{geq 2}+2geq 2+2=4$$ your expression is at least 12
    and equality is achieved if $a=b=c=1$.






    share|cite|improve this answer





















    • Oh this turned out to be too simple.I was trying to think in wrong direction .
      – iamredpanda
      1 hour ago










    • How do you show $x^2+1/x^2ge2$? Won't you need the same steps that can also show $x+1/xge2$?
      – Teepeemm
      34 mins ago










    • Exactly the same, just write $a=x^2$.
      – greedoid
      33 mins ago
















    5














    Since $$(x+1/x)^2 = underbrace{x^2+1/x^2}_{geq 2}+2geq 2+2=4$$ your expression is at least 12
    and equality is achieved if $a=b=c=1$.






    share|cite|improve this answer





















    • Oh this turned out to be too simple.I was trying to think in wrong direction .
      – iamredpanda
      1 hour ago










    • How do you show $x^2+1/x^2ge2$? Won't you need the same steps that can also show $x+1/xge2$?
      – Teepeemm
      34 mins ago










    • Exactly the same, just write $a=x^2$.
      – greedoid
      33 mins ago














    5












    5








    5






    Since $$(x+1/x)^2 = underbrace{x^2+1/x^2}_{geq 2}+2geq 2+2=4$$ your expression is at least 12
    and equality is achieved if $a=b=c=1$.






    share|cite|improve this answer












    Since $$(x+1/x)^2 = underbrace{x^2+1/x^2}_{geq 2}+2geq 2+2=4$$ your expression is at least 12
    and equality is achieved if $a=b=c=1$.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered 1 hour ago









    greedoid

    37.9k114794




    37.9k114794












    • Oh this turned out to be too simple.I was trying to think in wrong direction .
      – iamredpanda
      1 hour ago










    • How do you show $x^2+1/x^2ge2$? Won't you need the same steps that can also show $x+1/xge2$?
      – Teepeemm
      34 mins ago










    • Exactly the same, just write $a=x^2$.
      – greedoid
      33 mins ago


















    • Oh this turned out to be too simple.I was trying to think in wrong direction .
      – iamredpanda
      1 hour ago










    • How do you show $x^2+1/x^2ge2$? Won't you need the same steps that can also show $x+1/xge2$?
      – Teepeemm
      34 mins ago










    • Exactly the same, just write $a=x^2$.
      – greedoid
      33 mins ago
















    Oh this turned out to be too simple.I was trying to think in wrong direction .
    – iamredpanda
    1 hour ago




    Oh this turned out to be too simple.I was trying to think in wrong direction .
    – iamredpanda
    1 hour ago












    How do you show $x^2+1/x^2ge2$? Won't you need the same steps that can also show $x+1/xge2$?
    – Teepeemm
    34 mins ago




    How do you show $x^2+1/x^2ge2$? Won't you need the same steps that can also show $x+1/xge2$?
    – Teepeemm
    34 mins ago












    Exactly the same, just write $a=x^2$.
    – greedoid
    33 mins ago




    Exactly the same, just write $a=x^2$.
    – greedoid
    33 mins ago











    4














    Opimization is another way to solve this.



    $$f(a) = a+frac{1}{a} implies f’(a) = 1-a^{-2}$$



    Setting $f’(a) = 0$ to obtain the minimum, you have



    $$0 = 1-a^{-2} implies a^{-2} = 1 implies a = pm 1$$



    (For clarification, $a = -1$ gives the minimum absolute value for negative values of $a$.)



    Thus, the minimum is obtained when $a = pm 1$, so you have $left(a+frac{1}{a}right)^2 = (pm 2)^2 = 4$. The same goes for $b$ and $c$, so the minimum occurs when $a = b = c = pm 1$ and is $4+4+4 = 12$.






    share|cite|improve this answer























    • why does $f'(a)=0$ coincide with a minimum?
      – LinAlg
      1 hour ago










    • Because the stationary points occur where $f’(a) = 0$.
      – KM101
      1 hour ago






    • 2




      stationary points are not necessarily minima, and the function may not even be bounded below
      – LinAlg
      44 mins ago










    • I guess my reply wasn't clear enough. I didn't mean a stationary point always implies minima, but that it is the case in this example. For instance, for positive $a$, there is $frac{bpm n}{b}+frac{b}{bpm n} = frac{2left(b^2-bnright)+n^2}{b^2-bn}$, so there clearly is a minimum where $n = 0$, so $a = 1$.
      – KM101
      18 mins ago








    • 1




      @Tyberius I addressed that already. Since $a+frac{1}{a} < 0$ for $a < 0$, then $a = -1$ is a local maximum, and allows for the least absolute value for negative values of $a$.
      – KM101
      17 mins ago


















    4














    Opimization is another way to solve this.



    $$f(a) = a+frac{1}{a} implies f’(a) = 1-a^{-2}$$



    Setting $f’(a) = 0$ to obtain the minimum, you have



    $$0 = 1-a^{-2} implies a^{-2} = 1 implies a = pm 1$$



    (For clarification, $a = -1$ gives the minimum absolute value for negative values of $a$.)



    Thus, the minimum is obtained when $a = pm 1$, so you have $left(a+frac{1}{a}right)^2 = (pm 2)^2 = 4$. The same goes for $b$ and $c$, so the minimum occurs when $a = b = c = pm 1$ and is $4+4+4 = 12$.






    share|cite|improve this answer























    • why does $f'(a)=0$ coincide with a minimum?
      – LinAlg
      1 hour ago










    • Because the stationary points occur where $f’(a) = 0$.
      – KM101
      1 hour ago






    • 2




      stationary points are not necessarily minima, and the function may not even be bounded below
      – LinAlg
      44 mins ago










    • I guess my reply wasn't clear enough. I didn't mean a stationary point always implies minima, but that it is the case in this example. For instance, for positive $a$, there is $frac{bpm n}{b}+frac{b}{bpm n} = frac{2left(b^2-bnright)+n^2}{b^2-bn}$, so there clearly is a minimum where $n = 0$, so $a = 1$.
      – KM101
      18 mins ago








    • 1




      @Tyberius I addressed that already. Since $a+frac{1}{a} < 0$ for $a < 0$, then $a = -1$ is a local maximum, and allows for the least absolute value for negative values of $a$.
      – KM101
      17 mins ago
















    4












    4








    4






    Opimization is another way to solve this.



    $$f(a) = a+frac{1}{a} implies f’(a) = 1-a^{-2}$$



    Setting $f’(a) = 0$ to obtain the minimum, you have



    $$0 = 1-a^{-2} implies a^{-2} = 1 implies a = pm 1$$



    (For clarification, $a = -1$ gives the minimum absolute value for negative values of $a$.)



    Thus, the minimum is obtained when $a = pm 1$, so you have $left(a+frac{1}{a}right)^2 = (pm 2)^2 = 4$. The same goes for $b$ and $c$, so the minimum occurs when $a = b = c = pm 1$ and is $4+4+4 = 12$.






    share|cite|improve this answer














    Opimization is another way to solve this.



    $$f(a) = a+frac{1}{a} implies f’(a) = 1-a^{-2}$$



    Setting $f’(a) = 0$ to obtain the minimum, you have



    $$0 = 1-a^{-2} implies a^{-2} = 1 implies a = pm 1$$



    (For clarification, $a = -1$ gives the minimum absolute value for negative values of $a$.)



    Thus, the minimum is obtained when $a = pm 1$, so you have $left(a+frac{1}{a}right)^2 = (pm 2)^2 = 4$. The same goes for $b$ and $c$, so the minimum occurs when $a = b = c = pm 1$ and is $4+4+4 = 12$.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited 1 hour ago

























    answered 1 hour ago









    KM101

    4,886421




    4,886421












    • why does $f'(a)=0$ coincide with a minimum?
      – LinAlg
      1 hour ago










    • Because the stationary points occur where $f’(a) = 0$.
      – KM101
      1 hour ago






    • 2




      stationary points are not necessarily minima, and the function may not even be bounded below
      – LinAlg
      44 mins ago










    • I guess my reply wasn't clear enough. I didn't mean a stationary point always implies minima, but that it is the case in this example. For instance, for positive $a$, there is $frac{bpm n}{b}+frac{b}{bpm n} = frac{2left(b^2-bnright)+n^2}{b^2-bn}$, so there clearly is a minimum where $n = 0$, so $a = 1$.
      – KM101
      18 mins ago








    • 1




      @Tyberius I addressed that already. Since $a+frac{1}{a} < 0$ for $a < 0$, then $a = -1$ is a local maximum, and allows for the least absolute value for negative values of $a$.
      – KM101
      17 mins ago




















    • why does $f'(a)=0$ coincide with a minimum?
      – LinAlg
      1 hour ago










    • Because the stationary points occur where $f’(a) = 0$.
      – KM101
      1 hour ago






    • 2




      stationary points are not necessarily minima, and the function may not even be bounded below
      – LinAlg
      44 mins ago










    • I guess my reply wasn't clear enough. I didn't mean a stationary point always implies minima, but that it is the case in this example. For instance, for positive $a$, there is $frac{bpm n}{b}+frac{b}{bpm n} = frac{2left(b^2-bnright)+n^2}{b^2-bn}$, so there clearly is a minimum where $n = 0$, so $a = 1$.
      – KM101
      18 mins ago








    • 1




      @Tyberius I addressed that already. Since $a+frac{1}{a} < 0$ for $a < 0$, then $a = -1$ is a local maximum, and allows for the least absolute value for negative values of $a$.
      – KM101
      17 mins ago


















    why does $f'(a)=0$ coincide with a minimum?
    – LinAlg
    1 hour ago




    why does $f'(a)=0$ coincide with a minimum?
    – LinAlg
    1 hour ago












    Because the stationary points occur where $f’(a) = 0$.
    – KM101
    1 hour ago




    Because the stationary points occur where $f’(a) = 0$.
    – KM101
    1 hour ago




    2




    2




    stationary points are not necessarily minima, and the function may not even be bounded below
    – LinAlg
    44 mins ago




    stationary points are not necessarily minima, and the function may not even be bounded below
    – LinAlg
    44 mins ago












    I guess my reply wasn't clear enough. I didn't mean a stationary point always implies minima, but that it is the case in this example. For instance, for positive $a$, there is $frac{bpm n}{b}+frac{b}{bpm n} = frac{2left(b^2-bnright)+n^2}{b^2-bn}$, so there clearly is a minimum where $n = 0$, so $a = 1$.
    – KM101
    18 mins ago






    I guess my reply wasn't clear enough. I didn't mean a stationary point always implies minima, but that it is the case in this example. For instance, for positive $a$, there is $frac{bpm n}{b}+frac{b}{bpm n} = frac{2left(b^2-bnright)+n^2}{b^2-bn}$, so there clearly is a minimum where $n = 0$, so $a = 1$.
    – KM101
    18 mins ago






    1




    1




    @Tyberius I addressed that already. Since $a+frac{1}{a} < 0$ for $a < 0$, then $a = -1$ is a local maximum, and allows for the least absolute value for negative values of $a$.
    – KM101
    17 mins ago






    @Tyberius I addressed that already. Since $a+frac{1}{a} < 0$ for $a < 0$, then $a = -1$ is a local maximum, and allows for the least absolute value for negative values of $a$.
    – KM101
    17 mins ago













    0














    The given expression equals
    begin{align*}
    a^2 + b^2 + c^2 &+ frac{1}{a^2} + frac{1}{b^2} + frac{1}{c^2} + 6 \
    &geq 6 left(a^2 cdot b^2 cdot c^2 cdot frac{1}{a^2}frac{1}{b^2}frac{1}{c^2} right)^{frac{1}{6}} + 6 \
    &= 12
    end{align*}
    where we have used $AM - GM$ inequality.






    share|cite|improve this answer


























      0














      The given expression equals
      begin{align*}
      a^2 + b^2 + c^2 &+ frac{1}{a^2} + frac{1}{b^2} + frac{1}{c^2} + 6 \
      &geq 6 left(a^2 cdot b^2 cdot c^2 cdot frac{1}{a^2}frac{1}{b^2}frac{1}{c^2} right)^{frac{1}{6}} + 6 \
      &= 12
      end{align*}
      where we have used $AM - GM$ inequality.






      share|cite|improve this answer
























        0












        0








        0






        The given expression equals
        begin{align*}
        a^2 + b^2 + c^2 &+ frac{1}{a^2} + frac{1}{b^2} + frac{1}{c^2} + 6 \
        &geq 6 left(a^2 cdot b^2 cdot c^2 cdot frac{1}{a^2}frac{1}{b^2}frac{1}{c^2} right)^{frac{1}{6}} + 6 \
        &= 12
        end{align*}
        where we have used $AM - GM$ inequality.






        share|cite|improve this answer












        The given expression equals
        begin{align*}
        a^2 + b^2 + c^2 &+ frac{1}{a^2} + frac{1}{b^2} + frac{1}{c^2} + 6 \
        &geq 6 left(a^2 cdot b^2 cdot c^2 cdot frac{1}{a^2}frac{1}{b^2}frac{1}{c^2} right)^{frac{1}{6}} + 6 \
        &= 12
        end{align*}
        where we have used $AM - GM$ inequality.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 1 hour ago









        Muralidharan

        47516




        47516






























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            Plaza Victoria

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            Plaza Victoria Plaza Victoria en 2017 Otros nombres Plaza de La Victoria Localización El Almendral, Valparaíso, Chile Vías adyacentes Chacabuco, Molina, Plaza Victoria y Edwards Creación 1841 Metro Bellavista Ubicación 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 Coordenadas: 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 [editar datos en Wikidata] La Plaza Victoria , también llamada Plaza de la Victoria , [ 1 ] ​ es una plaza ubicada en el sector El Almendral del plan de la ciudad de Valparaíso, Chile, de importancia patrimonial, que incluye valiosas estatuas y especies vegetales. [ 2 ] ​ La plaza se creó como tal en la primera mitad del siglo XIX, tomando previamente los nombres de Plaza Nueva y Plaza de Orrego . Si bien desde sus inicios tuvo relevancia histórica, no fue sino a partir de fines de los años 1860 cuando comenzó a cobrar mayor importancia y se convirtió en uno de los principales centros sociales de l...
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            Brian Clough

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            Brian Clough Datos personales Nombre completo Brian Howard Clough Apodo(s) Old Big 'Ead, Cloughie [ 1 ] ​ Nacimiento Middlesbrough, Inglaterra, Reino Unido 21 de marzo de 1935 Nacionalidad(es) Fallecimiento Derby, Reino Unido 20 de septiembre de 2004 (69 años) Altura 1,78 [ 1 ] ​ metros Carrera Deporte Fútbol Debut deportivo 1955 (Como jugador) 1965 (Como entrenador) (Middlesbrough F. C. (Como jugador) Hartlepool United F. C. (Como entrenador) ) Posición Delantero Goles en clubes 269 [ 2 ] ​ Retirada deportiva 1964 (Como jugador) 1993 (Como entrenador) (Sunderland A. F. C. (Como jugador) Nottingham Forest (Como entrenador) ) Carrera internacional Part. 2 [editar datos en Wikidata] Brian Howard Clough , OBE [ 3 ] ​ (Middlesbrough, Inglaterra, Reino Unido, 21 de marzo de 1935 – Derby, Inglaterra, Reino Unido, 20 de septiembre de 2004) fue un jugador y entrenador británico de fútbol. Es conocido por...
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            Cáceres

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            Para otros usos de este término, véase Cáceres (desambiguación). Cáceres Municipio y ciudad Bandera Escudo Collage de Cáceres Cáceres Ubicación de Cáceres en España. Cáceres Ubicación de Cáceres en la provincia de Cáceres. País  España • Com. autónoma  Extremadura • Provincia  Cáceres • Partido judicial Cáceres Ubicación 39°28′23″N 6°22′16″O  /  39.473055555556, -6.3711111111111 Coordenadas: 39°28′23″N 6°22′16″O  /  39.473055555556, -6.3711111111111 • Altitud 457 [ 1 ] ​ msnm (mín.:211 [ 2 ] ​, máx.:708 [ 2 ] ​) Superficie 1750,33 km² Núcleos de población Cáceres Estación Arroyo-Malpartida Rincón de Ballesteros Valdesalor Fundación 34 a. C. Población 96 068 hab. (2018) • Densidad 54,8 hab./km² Gentilicio Cacereño/a [ 3 ] ​ Código postal 10001-10005, 10195 Alcaldesa (2015) María Elena Nevado del Campo [ 4 ] ​ Presupuesto 69.045.004€ [ 5 ] ​  (año ...
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