Proving that the arithmetic and geometric means of a collection of non-negative numbers lies between their...












2












$begingroup$



Consider non-negative real numbers $a_1, a_2, a_3, ... , a_n$. How can I prove that both the arithmetic mean (AM) and the geometric mean (GM) of $a_1, a_2, a_3, ... , a_n$ are contained in the interval $[x, y]$, where $x = text{minimum of} (a_1, a_2, a_3, ... , a_n)$ and $y = text{maximum of} (a_1, a_2, a_3, ... , a_n)$?




I know that the GM AM inequality states that GM $leq$ AM, so it would suffice to prove GM $geq$ x and AM $leq$ y. Am I correct so far, and if so, how should I proceed with the proof? Any hints or help in this direction would be greatly appreciated.



Thank you!






real-analysis proof-writing a.m.-g.m.-inequality







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$endgroup$












  • $begingroup$
    $a_1le y$, $a_2le y,ldots$ so $a_1a_2cdots a_nle y^n$. Does that help you show that the GM is $le y$?
    $endgroup$
    – Lord Shark the Unknown
    2 days ago
















2












$begingroup$



Consider non-negative real numbers $a_1, a_2, a_3, ... , a_n$. How can I prove that both the arithmetic mean (AM) and the geometric mean (GM) of $a_1, a_2, a_3, ... , a_n$ are contained in the interval $[x, y]$, where $x = text{minimum of} (a_1, a_2, a_3, ... , a_n)$ and $y = text{maximum of} (a_1, a_2, a_3, ... , a_n)$?




I know that the GM AM inequality states that GM $leq$ AM, so it would suffice to prove GM $geq$ x and AM $leq$ y. Am I correct so far, and if so, how should I proceed with the proof? Any hints or help in this direction would be greatly appreciated.



Thank you!






real-analysis proof-writing a.m.-g.m.-inequality







share|cite|improve this question











$endgroup$












  • $begingroup$
    $a_1le y$, $a_2le y,ldots$ so $a_1a_2cdots a_nle y^n$. Does that help you show that the GM is $le y$?
    $endgroup$
    – Lord Shark the Unknown
    2 days ago














2












2








2





$begingroup$



Consider non-negative real numbers $a_1, a_2, a_3, ... , a_n$. How can I prove that both the arithmetic mean (AM) and the geometric mean (GM) of $a_1, a_2, a_3, ... , a_n$ are contained in the interval $[x, y]$, where $x = text{minimum of} (a_1, a_2, a_3, ... , a_n)$ and $y = text{maximum of} (a_1, a_2, a_3, ... , a_n)$?




I know that the GM AM inequality states that GM $leq$ AM, so it would suffice to prove GM $geq$ x and AM $leq$ y. Am I correct so far, and if so, how should I proceed with the proof? Any hints or help in this direction would be greatly appreciated.



Thank you!






real-analysis proof-writing a.m.-g.m.-inequality







share|cite|improve this question











$endgroup$





Consider non-negative real numbers $a_1, a_2, a_3, ... , a_n$. How can I prove that both the arithmetic mean (AM) and the geometric mean (GM) of $a_1, a_2, a_3, ... , a_n$ are contained in the interval $[x, y]$, where $x = text{minimum of} (a_1, a_2, a_3, ... , a_n)$ and $y = text{maximum of} (a_1, a_2, a_3, ... , a_n)$?




I know that the GM AM inequality states that GM $leq$ AM, so it would suffice to prove GM $geq$ x and AM $leq$ y. Am I correct so far, and if so, how should I proceed with the proof? Any hints or help in this direction would be greatly appreciated.



Thank you!






real-analysis proof-writing a.m.-g.m.-inequality




real-analysis proof-writing a.m.-g.m.-inequality






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edited 2 days ago









Blue

47.7k870151




47.7k870151










asked 2 days ago









Zen'zZen'z

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183












  • $begingroup$
    $a_1le y$, $a_2le y,ldots$ so $a_1a_2cdots a_nle y^n$. Does that help you show that the GM is $le y$?
    $endgroup$
    – Lord Shark the Unknown
    2 days ago


















  • $begingroup$
    $a_1le y$, $a_2le y,ldots$ so $a_1a_2cdots a_nle y^n$. Does that help you show that the GM is $le y$?
    $endgroup$
    – Lord Shark the Unknown
    2 days ago
















$begingroup$
$a_1le y$, $a_2le y,ldots$ so $a_1a_2cdots a_nle y^n$. Does that help you show that the GM is $le y$?
$endgroup$
– Lord Shark the Unknown
2 days ago




$begingroup$
$a_1le y$, $a_2le y,ldots$ so $a_1a_2cdots a_nle y^n$. Does that help you show that the GM is $le y$?
$endgroup$
– Lord Shark the Unknown
2 days ago










3 Answers
3






active

oldest

votes


















4












$begingroup$

Yes, you are correct, it suffices to show that



i) $GMgeq x$, that is
$$a_1 a_2 a_3 cdots a_ngeq xcdot xcdot xcdots x= x^n$$
which holds because $a_kgeq x=min(a_1,a_2,a_3,dots,a_n)geq 0$ for $k=1,2,3,dots,n$.



ii) $AMleq y$, that is
$$a_1+a_2+a_3 +dots +a_nleq y+ y+ y+dots+ y=ny$$
which holds because $a_kleq y=max(a_1,a_2,a_3,dots,a_n)$ for $k=1,2,3,dots,n$.






share|cite|improve this answer











$endgroup$





















    4












    $begingroup$

    Just note that




    • $0leq x leq a_i Rightarrow sqrt[n]{x^n}leq sqrt[n]{a_1 cdot ldots cdot a_n}$

    • $a_i leq y Rightarrow frac{a_1 + cdots + a_n}{n}leq frac{y + cdots + y}{n} = y$






    share|cite|improve this answer









    $endgroup$





















      2












      $begingroup$

      GM: $sqrt[n]{a_1a_2cdots a_n}geqsqrt[n]{x^n}= x$



      AM: $frac{a_1+a_2+cdots+ a_n}{n}leq frac{ny}{n}=y$



      $ygeq(AM,GM)geq x$






      share|cite|improve this answer









      $endgroup$













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        3 Answers
        3






        active

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        3 Answers
        3






        active

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        active

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        active

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        4












        $begingroup$

        Yes, you are correct, it suffices to show that



        i) $GMgeq x$, that is
        $$a_1 a_2 a_3 cdots a_ngeq xcdot xcdot xcdots x= x^n$$
        which holds because $a_kgeq x=min(a_1,a_2,a_3,dots,a_n)geq 0$ for $k=1,2,3,dots,n$.



        ii) $AMleq y$, that is
        $$a_1+a_2+a_3 +dots +a_nleq y+ y+ y+dots+ y=ny$$
        which holds because $a_kleq y=max(a_1,a_2,a_3,dots,a_n)$ for $k=1,2,3,dots,n$.






        share|cite|improve this answer











        $endgroup$


















          4












          $begingroup$

          Yes, you are correct, it suffices to show that



          i) $GMgeq x$, that is
          $$a_1 a_2 a_3 cdots a_ngeq xcdot xcdot xcdots x= x^n$$
          which holds because $a_kgeq x=min(a_1,a_2,a_3,dots,a_n)geq 0$ for $k=1,2,3,dots,n$.



          ii) $AMleq y$, that is
          $$a_1+a_2+a_3 +dots +a_nleq y+ y+ y+dots+ y=ny$$
          which holds because $a_kleq y=max(a_1,a_2,a_3,dots,a_n)$ for $k=1,2,3,dots,n$.






          share|cite|improve this answer











          $endgroup$
















            4












            4








            4





            $begingroup$

            Yes, you are correct, it suffices to show that



            i) $GMgeq x$, that is
            $$a_1 a_2 a_3 cdots a_ngeq xcdot xcdot xcdots x= x^n$$
            which holds because $a_kgeq x=min(a_1,a_2,a_3,dots,a_n)geq 0$ for $k=1,2,3,dots,n$.



            ii) $AMleq y$, that is
            $$a_1+a_2+a_3 +dots +a_nleq y+ y+ y+dots+ y=ny$$
            which holds because $a_kleq y=max(a_1,a_2,a_3,dots,a_n)$ for $k=1,2,3,dots,n$.






            share|cite|improve this answer











            $endgroup$



            Yes, you are correct, it suffices to show that



            i) $GMgeq x$, that is
            $$a_1 a_2 a_3 cdots a_ngeq xcdot xcdot xcdots x= x^n$$
            which holds because $a_kgeq x=min(a_1,a_2,a_3,dots,a_n)geq 0$ for $k=1,2,3,dots,n$.



            ii) $AMleq y$, that is
            $$a_1+a_2+a_3 +dots +a_nleq y+ y+ y+dots+ y=ny$$
            which holds because $a_kleq y=max(a_1,a_2,a_3,dots,a_n)$ for $k=1,2,3,dots,n$.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited 2 days ago

























            answered 2 days ago









            Robert ZRobert Z

            94.5k1062133




            94.5k1062133























                4












                $begingroup$

                Just note that




                • $0leq x leq a_i Rightarrow sqrt[n]{x^n}leq sqrt[n]{a_1 cdot ldots cdot a_n}$

                • $a_i leq y Rightarrow frac{a_1 + cdots + a_n}{n}leq frac{y + cdots + y}{n} = y$






                share|cite|improve this answer









                $endgroup$


















                  4












                  $begingroup$

                  Just note that




                  • $0leq x leq a_i Rightarrow sqrt[n]{x^n}leq sqrt[n]{a_1 cdot ldots cdot a_n}$

                  • $a_i leq y Rightarrow frac{a_1 + cdots + a_n}{n}leq frac{y + cdots + y}{n} = y$






                  share|cite|improve this answer









                  $endgroup$
















                    4












                    4








                    4





                    $begingroup$

                    Just note that




                    • $0leq x leq a_i Rightarrow sqrt[n]{x^n}leq sqrt[n]{a_1 cdot ldots cdot a_n}$

                    • $a_i leq y Rightarrow frac{a_1 + cdots + a_n}{n}leq frac{y + cdots + y}{n} = y$






                    share|cite|improve this answer









                    $endgroup$



                    Just note that




                    • $0leq x leq a_i Rightarrow sqrt[n]{x^n}leq sqrt[n]{a_1 cdot ldots cdot a_n}$

                    • $a_i leq y Rightarrow frac{a_1 + cdots + a_n}{n}leq frac{y + cdots + y}{n} = y$







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered 2 days ago









                    trancelocationtrancelocation

                    9,6801722




                    9,6801722























                        2












                        $begingroup$

                        GM: $sqrt[n]{a_1a_2cdots a_n}geqsqrt[n]{x^n}= x$



                        AM: $frac{a_1+a_2+cdots+ a_n}{n}leq frac{ny}{n}=y$



                        $ygeq(AM,GM)geq x$






                        share|cite|improve this answer









                        $endgroup$


















                          2












                          $begingroup$

                          GM: $sqrt[n]{a_1a_2cdots a_n}geqsqrt[n]{x^n}= x$



                          AM: $frac{a_1+a_2+cdots+ a_n}{n}leq frac{ny}{n}=y$



                          $ygeq(AM,GM)geq x$






                          share|cite|improve this answer









                          $endgroup$
















                            2












                            2








                            2





                            $begingroup$

                            GM: $sqrt[n]{a_1a_2cdots a_n}geqsqrt[n]{x^n}= x$



                            AM: $frac{a_1+a_2+cdots+ a_n}{n}leq frac{ny}{n}=y$



                            $ygeq(AM,GM)geq x$






                            share|cite|improve this answer









                            $endgroup$



                            GM: $sqrt[n]{a_1a_2cdots a_n}geqsqrt[n]{x^n}= x$



                            AM: $frac{a_1+a_2+cdots+ a_n}{n}leq frac{ny}{n}=y$



                            $ygeq(AM,GM)geq x$







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered 2 days ago









                            LeeLee

                            234111




                            234111






























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