Inequality for sum of squares












0












$begingroup$


let $a_t$ and $b_t$ be any real number. Is the following inequality true?
begin{equation}
frac{1}{T} sum_{t=1}^{T}2a_t^2 + frac{1}{T} sum_{t=1}^{T}2b_t^2 > frac{1}{T} sum_{t=1}^{T}(a_t + b_t)^2
end{equation}
Any comments is appreciated.






inequality







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  • 1




    $begingroup$
    Can you say something about the variables?
    $endgroup$
    – Dr. Sonnhard Graubner
    Jan 19 at 18:11










  • $begingroup$
    Can you do the $T=1$ case?
    $endgroup$
    – Lord Shark the Unknown
    Jan 19 at 18:15










  • $begingroup$
    For the case $T=1$, what about $a_1=b_1$? You would have to prove $$2a_1^2+2b_1^2>(a_1+b_1)^2$$ $$iff 4a_1^2>(2a_1)^2=4a_1^2$$ which is obvioulsy false... You might have to prove that $$begin{equation} frac{1}{T} sum_{t=1}^{T}2a_t^2 + frac{1}{T} sum_{t=1}^{T}2b_t^2 mathbf{≥} frac{1}{T} sum_{t=1}^{T}(a_t + b_t)^2 end{equation}$$
    $endgroup$
    – Dr. Mathva
    Jan 19 at 18:19


















0












$begingroup$


let $a_t$ and $b_t$ be any real number. Is the following inequality true?
begin{equation}
frac{1}{T} sum_{t=1}^{T}2a_t^2 + frac{1}{T} sum_{t=1}^{T}2b_t^2 > frac{1}{T} sum_{t=1}^{T}(a_t + b_t)^2
end{equation}
Any comments is appreciated.






inequality







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Can you say something about the variables?
    $endgroup$
    – Dr. Sonnhard Graubner
    Jan 19 at 18:11










  • $begingroup$
    Can you do the $T=1$ case?
    $endgroup$
    – Lord Shark the Unknown
    Jan 19 at 18:15










  • $begingroup$
    For the case $T=1$, what about $a_1=b_1$? You would have to prove $$2a_1^2+2b_1^2>(a_1+b_1)^2$$ $$iff 4a_1^2>(2a_1)^2=4a_1^2$$ which is obvioulsy false... You might have to prove that $$begin{equation} frac{1}{T} sum_{t=1}^{T}2a_t^2 + frac{1}{T} sum_{t=1}^{T}2b_t^2 mathbf{≥} frac{1}{T} sum_{t=1}^{T}(a_t + b_t)^2 end{equation}$$
    $endgroup$
    – Dr. Mathva
    Jan 19 at 18:19
















0












0








0





$begingroup$


let $a_t$ and $b_t$ be any real number. Is the following inequality true?
begin{equation}
frac{1}{T} sum_{t=1}^{T}2a_t^2 + frac{1}{T} sum_{t=1}^{T}2b_t^2 > frac{1}{T} sum_{t=1}^{T}(a_t + b_t)^2
end{equation}
Any comments is appreciated.






inequality







share|cite|improve this question









$endgroup$




let $a_t$ and $b_t$ be any real number. Is the following inequality true?
begin{equation}
frac{1}{T} sum_{t=1}^{T}2a_t^2 + frac{1}{T} sum_{t=1}^{T}2b_t^2 > frac{1}{T} sum_{t=1}^{T}(a_t + b_t)^2
end{equation}
Any comments is appreciated.






inequality




inequality






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share|cite|improve this question











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asked Jan 19 at 18:09









Mike LMike L

1




1








  • 1




    $begingroup$
    Can you say something about the variables?
    $endgroup$
    – Dr. Sonnhard Graubner
    Jan 19 at 18:11










  • $begingroup$
    Can you do the $T=1$ case?
    $endgroup$
    – Lord Shark the Unknown
    Jan 19 at 18:15










  • $begingroup$
    For the case $T=1$, what about $a_1=b_1$? You would have to prove $$2a_1^2+2b_1^2>(a_1+b_1)^2$$ $$iff 4a_1^2>(2a_1)^2=4a_1^2$$ which is obvioulsy false... You might have to prove that $$begin{equation} frac{1}{T} sum_{t=1}^{T}2a_t^2 + frac{1}{T} sum_{t=1}^{T}2b_t^2 mathbf{≥} frac{1}{T} sum_{t=1}^{T}(a_t + b_t)^2 end{equation}$$
    $endgroup$
    – Dr. Mathva
    Jan 19 at 18:19
















  • 1




    $begingroup$
    Can you say something about the variables?
    $endgroup$
    – Dr. Sonnhard Graubner
    Jan 19 at 18:11










  • $begingroup$
    Can you do the $T=1$ case?
    $endgroup$
    – Lord Shark the Unknown
    Jan 19 at 18:15










  • $begingroup$
    For the case $T=1$, what about $a_1=b_1$? You would have to prove $$2a_1^2+2b_1^2>(a_1+b_1)^2$$ $$iff 4a_1^2>(2a_1)^2=4a_1^2$$ which is obvioulsy false... You might have to prove that $$begin{equation} frac{1}{T} sum_{t=1}^{T}2a_t^2 + frac{1}{T} sum_{t=1}^{T}2b_t^2 mathbf{≥} frac{1}{T} sum_{t=1}^{T}(a_t + b_t)^2 end{equation}$$
    $endgroup$
    – Dr. Mathva
    Jan 19 at 18:19










1




1




$begingroup$
Can you say something about the variables?
$endgroup$
– Dr. Sonnhard Graubner
Jan 19 at 18:11




$begingroup$
Can you say something about the variables?
$endgroup$
– Dr. Sonnhard Graubner
Jan 19 at 18:11












$begingroup$
Can you do the $T=1$ case?
$endgroup$
– Lord Shark the Unknown
Jan 19 at 18:15




$begingroup$
Can you do the $T=1$ case?
$endgroup$
– Lord Shark the Unknown
Jan 19 at 18:15












$begingroup$
For the case $T=1$, what about $a_1=b_1$? You would have to prove $$2a_1^2+2b_1^2>(a_1+b_1)^2$$ $$iff 4a_1^2>(2a_1)^2=4a_1^2$$ which is obvioulsy false... You might have to prove that $$begin{equation} frac{1}{T} sum_{t=1}^{T}2a_t^2 + frac{1}{T} sum_{t=1}^{T}2b_t^2 mathbf{≥} frac{1}{T} sum_{t=1}^{T}(a_t + b_t)^2 end{equation}$$
$endgroup$
– Dr. Mathva
Jan 19 at 18:19






$begingroup$
For the case $T=1$, what about $a_1=b_1$? You would have to prove $$2a_1^2+2b_1^2>(a_1+b_1)^2$$ $$iff 4a_1^2>(2a_1)^2=4a_1^2$$ which is obvioulsy false... You might have to prove that $$begin{equation} frac{1}{T} sum_{t=1}^{T}2a_t^2 + frac{1}{T} sum_{t=1}^{T}2b_t^2 mathbf{≥} frac{1}{T} sum_{t=1}^{T}(a_t + b_t)^2 end{equation}$$
$endgroup$
– Dr. Mathva
Jan 19 at 18:19












2 Answers
2






active

oldest

votes


















1












$begingroup$

Just observe that



$2a_t^2+2b_t^2-(a_t+b_t)^2=(a_t-b_t)^2geq0$



That proves your inequality (if you exchange the "$>$" with a "$geq$").






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    As I showed in the comments, it doesn't have to be necessarily "bigger", so I'll prove the inequality
    $$frac{1}{T} sum_{t=1}^{T}2a_t^2 + frac{1}{T} sum_{t=1}^{T}2b_t^2 ≥ frac{1}{T} sum_{t=1}^{T}(a_t + b_t)^2;; a_t,b_tinmathbb R ;text{and} ;Tinmathbb N$$
    Note that
    $$frac{1}{T} sum_{t=1}^{T}2a_t^2 + frac{1}{T} sum_{t=1}^{T}2b_t^2 ≥ frac{1}{T} sum_{t=1}^{T}(a_t + b_t)^2ifffrac{1}{T}sum^T_{t=1}2a_t^2+2b_t^2≥frac{1}{T}sum_{t=1}^T(a_t + b_t)^2$$
    $$iff sum^T_{t=1}2a_t^2+2b_t^2≥sum_{t=1}^T(a_t + b_t)^2$$
    $$iff2a_t^2+2b_t^2≥(a_t+b_t)^2iff a_t^2+b_t^2≥2a_tb_tiff a_t^2+b_t^2-2a_tb_t≥0$$
    Which obviously holds for every $a_t,b_tinmathbb R$ since
    $$a_t^2+b_t^2-2a_tb_t=(a_t-b_t)^2≥0$$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Thank you very much Dr. Mathva!
      $endgroup$
      – Mike L
      Jan 20 at 9:17










    • $begingroup$
      @MikeL you can also upvote the answer if you liked it ;)
      $endgroup$
      – Dr. Mathva
      Jan 20 at 9:25











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    2 Answers
    2






    active

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    2 Answers
    2






    active

    oldest

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    active

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    active

    oldest

    votes









    1












    $begingroup$

    Just observe that



    $2a_t^2+2b_t^2-(a_t+b_t)^2=(a_t-b_t)^2geq0$



    That proves your inequality (if you exchange the "$>$" with a "$geq$").






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Just observe that



      $2a_t^2+2b_t^2-(a_t+b_t)^2=(a_t-b_t)^2geq0$



      That proves your inequality (if you exchange the "$>$" with a "$geq$").






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Just observe that



        $2a_t^2+2b_t^2-(a_t+b_t)^2=(a_t-b_t)^2geq0$



        That proves your inequality (if you exchange the "$>$" with a "$geq$").






        share|cite|improve this answer









        $endgroup$



        Just observe that



        $2a_t^2+2b_t^2-(a_t+b_t)^2=(a_t-b_t)^2geq0$



        That proves your inequality (if you exchange the "$>$" with a "$geq$").







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 19 at 18:28









        maxmilgrammaxmilgram

        6557




        6557























            0












            $begingroup$

            As I showed in the comments, it doesn't have to be necessarily "bigger", so I'll prove the inequality
            $$frac{1}{T} sum_{t=1}^{T}2a_t^2 + frac{1}{T} sum_{t=1}^{T}2b_t^2 ≥ frac{1}{T} sum_{t=1}^{T}(a_t + b_t)^2;; a_t,b_tinmathbb R ;text{and} ;Tinmathbb N$$
            Note that
            $$frac{1}{T} sum_{t=1}^{T}2a_t^2 + frac{1}{T} sum_{t=1}^{T}2b_t^2 ≥ frac{1}{T} sum_{t=1}^{T}(a_t + b_t)^2ifffrac{1}{T}sum^T_{t=1}2a_t^2+2b_t^2≥frac{1}{T}sum_{t=1}^T(a_t + b_t)^2$$
            $$iff sum^T_{t=1}2a_t^2+2b_t^2≥sum_{t=1}^T(a_t + b_t)^2$$
            $$iff2a_t^2+2b_t^2≥(a_t+b_t)^2iff a_t^2+b_t^2≥2a_tb_tiff a_t^2+b_t^2-2a_tb_t≥0$$
            Which obviously holds for every $a_t,b_tinmathbb R$ since
            $$a_t^2+b_t^2-2a_tb_t=(a_t-b_t)^2≥0$$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Thank you very much Dr. Mathva!
              $endgroup$
              – Mike L
              Jan 20 at 9:17










            • $begingroup$
              @MikeL you can also upvote the answer if you liked it ;)
              $endgroup$
              – Dr. Mathva
              Jan 20 at 9:25
















            0












            $begingroup$

            As I showed in the comments, it doesn't have to be necessarily "bigger", so I'll prove the inequality
            $$frac{1}{T} sum_{t=1}^{T}2a_t^2 + frac{1}{T} sum_{t=1}^{T}2b_t^2 ≥ frac{1}{T} sum_{t=1}^{T}(a_t + b_t)^2;; a_t,b_tinmathbb R ;text{and} ;Tinmathbb N$$
            Note that
            $$frac{1}{T} sum_{t=1}^{T}2a_t^2 + frac{1}{T} sum_{t=1}^{T}2b_t^2 ≥ frac{1}{T} sum_{t=1}^{T}(a_t + b_t)^2ifffrac{1}{T}sum^T_{t=1}2a_t^2+2b_t^2≥frac{1}{T}sum_{t=1}^T(a_t + b_t)^2$$
            $$iff sum^T_{t=1}2a_t^2+2b_t^2≥sum_{t=1}^T(a_t + b_t)^2$$
            $$iff2a_t^2+2b_t^2≥(a_t+b_t)^2iff a_t^2+b_t^2≥2a_tb_tiff a_t^2+b_t^2-2a_tb_t≥0$$
            Which obviously holds for every $a_t,b_tinmathbb R$ since
            $$a_t^2+b_t^2-2a_tb_t=(a_t-b_t)^2≥0$$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Thank you very much Dr. Mathva!
              $endgroup$
              – Mike L
              Jan 20 at 9:17










            • $begingroup$
              @MikeL you can also upvote the answer if you liked it ;)
              $endgroup$
              – Dr. Mathva
              Jan 20 at 9:25














            0












            0








            0





            $begingroup$

            As I showed in the comments, it doesn't have to be necessarily "bigger", so I'll prove the inequality
            $$frac{1}{T} sum_{t=1}^{T}2a_t^2 + frac{1}{T} sum_{t=1}^{T}2b_t^2 ≥ frac{1}{T} sum_{t=1}^{T}(a_t + b_t)^2;; a_t,b_tinmathbb R ;text{and} ;Tinmathbb N$$
            Note that
            $$frac{1}{T} sum_{t=1}^{T}2a_t^2 + frac{1}{T} sum_{t=1}^{T}2b_t^2 ≥ frac{1}{T} sum_{t=1}^{T}(a_t + b_t)^2ifffrac{1}{T}sum^T_{t=1}2a_t^2+2b_t^2≥frac{1}{T}sum_{t=1}^T(a_t + b_t)^2$$
            $$iff sum^T_{t=1}2a_t^2+2b_t^2≥sum_{t=1}^T(a_t + b_t)^2$$
            $$iff2a_t^2+2b_t^2≥(a_t+b_t)^2iff a_t^2+b_t^2≥2a_tb_tiff a_t^2+b_t^2-2a_tb_t≥0$$
            Which obviously holds for every $a_t,b_tinmathbb R$ since
            $$a_t^2+b_t^2-2a_tb_t=(a_t-b_t)^2≥0$$






            share|cite|improve this answer











            $endgroup$



            As I showed in the comments, it doesn't have to be necessarily "bigger", so I'll prove the inequality
            $$frac{1}{T} sum_{t=1}^{T}2a_t^2 + frac{1}{T} sum_{t=1}^{T}2b_t^2 ≥ frac{1}{T} sum_{t=1}^{T}(a_t + b_t)^2;; a_t,b_tinmathbb R ;text{and} ;Tinmathbb N$$
            Note that
            $$frac{1}{T} sum_{t=1}^{T}2a_t^2 + frac{1}{T} sum_{t=1}^{T}2b_t^2 ≥ frac{1}{T} sum_{t=1}^{T}(a_t + b_t)^2ifffrac{1}{T}sum^T_{t=1}2a_t^2+2b_t^2≥frac{1}{T}sum_{t=1}^T(a_t + b_t)^2$$
            $$iff sum^T_{t=1}2a_t^2+2b_t^2≥sum_{t=1}^T(a_t + b_t)^2$$
            $$iff2a_t^2+2b_t^2≥(a_t+b_t)^2iff a_t^2+b_t^2≥2a_tb_tiff a_t^2+b_t^2-2a_tb_t≥0$$
            Which obviously holds for every $a_t,b_tinmathbb R$ since
            $$a_t^2+b_t^2-2a_tb_t=(a_t-b_t)^2≥0$$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Jan 19 at 19:22

























            answered Jan 19 at 18:32









            Dr. MathvaDr. Mathva

            2,016324




            2,016324












            • $begingroup$
              Thank you very much Dr. Mathva!
              $endgroup$
              – Mike L
              Jan 20 at 9:17










            • $begingroup$
              @MikeL you can also upvote the answer if you liked it ;)
              $endgroup$
              – Dr. Mathva
              Jan 20 at 9:25


















            • $begingroup$
              Thank you very much Dr. Mathva!
              $endgroup$
              – Mike L
              Jan 20 at 9:17










            • $begingroup$
              @MikeL you can also upvote the answer if you liked it ;)
              $endgroup$
              – Dr. Mathva
              Jan 20 at 9:25
















            $begingroup$
            Thank you very much Dr. Mathva!
            $endgroup$
            – Mike L
            Jan 20 at 9:17




            $begingroup$
            Thank you very much Dr. Mathva!
            $endgroup$
            – Mike L
            Jan 20 at 9:17












            $begingroup$
            @MikeL you can also upvote the answer if you liked it ;)
            $endgroup$
            – Dr. Mathva
            Jan 20 at 9:25




            $begingroup$
            @MikeL you can also upvote the answer if you liked it ;)
            $endgroup$
            – Dr. Mathva
            Jan 20 at 9:25


















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