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





















    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3058039%2ffinding-minimum-value-of-the-expression%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    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

        48516




        48516






























            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.





            Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


            Please pay close attention to the following guidance:


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3058039%2ffinding-minimum-value-of-the-expression%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            -

            Popular posts from this blog

            Understanding the information contained in the Deep Space Network XML data?

            Image
            up vote 1 down vote favorite NASA's Deep Space Network activity can be viewed conveniently at https://eyes.nasa.gov/dsn/dsn.html but if you would like data in a more raw XML form, it can be accessed as described in this answer Below is a small Python script that reads the XML into a list of dictionaries, and saves it to disk as a JSON. The dictionaries are organized at the higher level by dish, then by spacecraft per dish. Question: What are the most-likely meanings of all the data field contained in the Deep Space Network XML data? This is the example url from the other answer, used in the script below: https://eyes.nasa.gov/dsn/data/1365107113.xml def dictify(r,root=True): # https://stackoverflow.com/questions/2148119/how-to-convert-an-xml-string-to-a-dictionary-in-python/30923963#30923963 if root: ret
            Read more

            Ross-on-Wye

            Image
            Ross-on-Wye Welsh: Rhosan ar Wy Town centre, looking north from Market House Ross-on-Wye Ross-on-Wye shown within Herefordshire Population 10,700 (2011) [1] OS grid reference SO597241 Unitary authority Herefordshire Ceremonial county Herefordshire Region West Midlands Country England Sovereign state United Kingdom Post town ROSS-ON-WYE Postcode district HR9 Dialling code 01989 Police West Mercia Fire Hereford and Worcester Ambulance West Midlands EU Parliament West Midlands UK Parliament Hereford and South Herefordshire List of places UK England Herefordshire 51°54′50″N 2°35′13″W  /  51.914°N 2.587°W  / 51.914; -2.587 Coordinates: 51°54′50″N 2°35′13″W  /  51.914°N 2.587°W  / 51.914; -2.587 Ross-on-Wye (Welsh: Rhosan ar Wy [2] ) is a small market town with a population of 10,700 (according to the 2011 census), in south eastern Herefordshire, England, located on the River Wye, and on
            Read more

            Eastern Orthodox Church

            Image
            Not to be confused with Oriental Orthodoxy, Eastern Christianity, or Eastern Catholic Churches. "Eastern Orthodoxy", "Orthodox Church", "Orthodox Catholic Church", and "Orthodox Christian Church" redirect here. For other uses, see Orthodox (disambiguation). This article may be too long to read and navigate comfortably . The readable prose size is 132 kilobytes. Please consider splitting content into sub-articles, condensing it, or adding or removing subheadings. ( January 2018 ) Part of a series on the Eastern Orthodox Church Mosaic of Christ Pantocrator, Hagia Sophia Overview Structure Theology (History of theology) Liturgy Church history Holy Mysteries View of salvation View of Mary View of icons Background Crucifixion / Resurrection / Ascension of Jesus Christianity Christian Church Apostolic succession Four Marks of the Church Orthodoxy Organization Autocephaly Patriarchate
            Read more