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
    7 hours ago










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












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










  • See also: math.stackexchange.com/questions/487486/…
    – lab bhattacharjee
    4 hours 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
    7 hours ago










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












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










  • See also: math.stackexchange.com/questions/487486/…
    – lab bhattacharjee
    4 hours ago














1












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 6 hours ago









greedoid

37.9k114794




37.9k114794










asked 7 hours ago









iamredpanda

123




123












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










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












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










  • See also: math.stackexchange.com/questions/487486/…
    – lab bhattacharjee
    4 hours ago


















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










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












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










  • See also: math.stackexchange.com/questions/487486/…
    – lab bhattacharjee
    4 hours ago
















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




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












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






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














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




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












See also: math.stackexchange.com/questions/487486/…
– lab bhattacharjee
4 hours ago




See also: math.stackexchange.com/questions/487486/…
– lab bhattacharjee
4 hours 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
    6 hours ago






  • 1




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










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










  • @Teepeemm , with squares we are sure the numbers are positive. It is not so for $x+1/x$ which is less than $-2$ if $x<0.$
    – user376343
    4 hours ago










  • Then what does your answer add? If the same steps can show $x+1/xge2$, then how does this answer help OP? @user376343 Squares being positive shows us that $x^2+1/x^2ge0$, not $ge2$.
    – Teepeemm
    20 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
    6 hours ago










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






  • 2




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





















1














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
      6 hours ago






    • 1




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










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










    • @Teepeemm , with squares we are sure the numbers are positive. It is not so for $x+1/x$ which is less than $-2$ if $x<0.$
      – user376343
      4 hours ago










    • Then what does your answer add? If the same steps can show $x+1/xge2$, then how does this answer help OP? @user376343 Squares being positive shows us that $x^2+1/x^2ge0$, not $ge2$.
      – Teepeemm
      20 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
      6 hours ago






    • 1




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










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










    • @Teepeemm , with squares we are sure the numbers are positive. It is not so for $x+1/x$ which is less than $-2$ if $x<0.$
      – user376343
      4 hours ago










    • Then what does your answer add? If the same steps can show $x+1/xge2$, then how does this answer help OP? @user376343 Squares being positive shows us that $x^2+1/x^2ge0$, not $ge2$.
      – Teepeemm
      20 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 6 hours ago









    greedoid

    37.9k114794




    37.9k114794












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






    • 1




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










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










    • @Teepeemm , with squares we are sure the numbers are positive. It is not so for $x+1/x$ which is less than $-2$ if $x<0.$
      – user376343
      4 hours ago










    • Then what does your answer add? If the same steps can show $x+1/xge2$, then how does this answer help OP? @user376343 Squares being positive shows us that $x^2+1/x^2ge0$, not $ge2$.
      – Teepeemm
      20 mins ago


















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






    • 1




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










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










    • @Teepeemm , with squares we are sure the numbers are positive. It is not so for $x+1/x$ which is less than $-2$ if $x<0.$
      – user376343
      4 hours ago










    • Then what does your answer add? If the same steps can show $x+1/xge2$, then how does this answer help OP? @user376343 Squares being positive shows us that $x^2+1/x^2ge0$, not $ge2$.
      – Teepeemm
      20 mins ago
















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




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




    1




    1




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












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




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












    @Teepeemm , with squares we are sure the numbers are positive. It is not so for $x+1/x$ which is less than $-2$ if $x<0.$
    – user376343
    4 hours ago




    @Teepeemm , with squares we are sure the numbers are positive. It is not so for $x+1/x$ which is less than $-2$ if $x<0.$
    – user376343
    4 hours ago












    Then what does your answer add? If the same steps can show $x+1/xge2$, then how does this answer help OP? @user376343 Squares being positive shows us that $x^2+1/x^2ge0$, not $ge2$.
    – Teepeemm
    20 mins ago




    Then what does your answer add? If the same steps can show $x+1/xge2$, then how does this answer help OP? @user376343 Squares being positive shows us that $x^2+1/x^2ge0$, not $ge2$.
    – Teepeemm
    20 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
      6 hours ago










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






    • 2




      stationary points are not necessarily minima, and the function may not even be bounded below
      – LinAlg
      5 hours 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
      5 hours 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
      6 hours ago










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






    • 2




      stationary points are not necessarily minima, and the function may not even be bounded below
      – LinAlg
      5 hours 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
      5 hours 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 6 hours ago

























    answered 6 hours ago









    KM101

    4,886421




    4,886421












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










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






    • 2




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




















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










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






    • 2




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


















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




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












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




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




    2




    2




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




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













    1














    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


























      1














      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
























        1












        1








        1






        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 6 hours ago









        Muralidharan

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