Find the maximum constant such that the inequality

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Let $a;b>0$. Find the maximum constant such that the inequality $$frac1a^2+b^2+frac1a^2+frac1b^2ge frac8+2kleft(a+bright)^2$$



Let $a=1$ then we have: $-frack-12a^2ge 0Leftrightarrow kle 1$. So we will prove $k=1$ is the maximum constant.



$$fracleft(a-bright)^2left(a^4+4a^3b+a^2b^2+4ab^3+b^4right)a^2b^2left(a+bright)^2left(a^2+b^2right)ge 0$$



Is that true ?




inequality optimization fractions maxima-minima a.m.-g.m.-inequality




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  • I think you are right.
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Let $a;b>0$. Find the maximum constant such that the inequality $$frac1a^2+b^2+frac1a^2+frac1b^2ge frac8+2kleft(a+bright)^2$$



Let $a=1$ then we have: $-frack-12a^2ge 0Leftrightarrow kle 1$. So we will prove $k=1$ is the maximum constant.



$$fracleft(a-bright)^2left(a^4+4a^3b+a^2b^2+4ab^3+b^4right)a^2b^2left(a+bright)^2left(a^2+b^2right)ge 0$$



Is that true ?




inequality optimization fractions maxima-minima a.m.-g.m.-inequality




share|cite|improve this question





















  • I think you are right.
    – Michael Rozenberg
    2 days ago












up vote
0
down vote

favorite









up vote
0
down vote

favorite











Let $a;b>0$. Find the maximum constant such that the inequality $$frac1a^2+b^2+frac1a^2+frac1b^2ge frac8+2kleft(a+bright)^2$$



Let $a=1$ then we have: $-frack-12a^2ge 0Leftrightarrow kle 1$. So we will prove $k=1$ is the maximum constant.



$$fracleft(a-bright)^2left(a^4+4a^3b+a^2b^2+4ab^3+b^4right)a^2b^2left(a+bright)^2left(a^2+b^2right)ge 0$$



Is that true ?




inequality optimization fractions maxima-minima a.m.-g.m.-inequality




share|cite|improve this question













Let $a;b>0$. Find the maximum constant such that the inequality $$frac1a^2+b^2+frac1a^2+frac1b^2ge frac8+2kleft(a+bright)^2$$



Let $a=1$ then we have: $-frack-12a^2ge 0Leftrightarrow kle 1$. So we will prove $k=1$ is the maximum constant.



$$fracleft(a-bright)^2left(a^4+4a^3b+a^2b^2+4ab^3+b^4right)a^2b^2left(a+bright)^2left(a^2+b^2right)ge 0$$



Is that true ?





inequality optimization fractions maxima-minima a.m.-g.m.-inequality





share|cite|improve this question












share|cite|improve this question




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edited 2 days ago









Michael Rozenberg

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asked 2 days ago









Nguyễn Duy Linh

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  • I think you are right.
    – Michael Rozenberg
    2 days ago
















  • I think you are right.
    – Michael Rozenberg
    2 days ago















I think you are right.
– Michael Rozenberg
2 days ago




I think you are right.
– Michael Rozenberg
2 days ago










1 Answer
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accepted










Your solution is right.



The proof of the final inequality we can make also by AM-GM.



Indeed,
$$frac1a^2+b^2+frac1a^2+frac1b^2=frac1a^2+b^2+fraca^2+b^2a^2b^2=$$
$$=frac1a^2+b^2+4cdotfraca^2+b^24a^2b^2geq5sqrt[5]frac1a^2+b^2left(fraca^2+b^24a^2b^2right)^4=$$
$$=5sqrt[5]frac(a^2+b^2)^3256a^8b^8geq5sqrt[5]frac(2ab)^3256a^8b^8=5sqrt[5]frac132a^5b^5=frac104abgeqfrac10(a+b)^2.$$






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    1 Answer
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    active

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    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    0
    down vote



    accepted










    Your solution is right.



    The proof of the final inequality we can make also by AM-GM.



    Indeed,
    $$frac1a^2+b^2+frac1a^2+frac1b^2=frac1a^2+b^2+fraca^2+b^2a^2b^2=$$
    $$=frac1a^2+b^2+4cdotfraca^2+b^24a^2b^2geq5sqrt[5]frac1a^2+b^2left(fraca^2+b^24a^2b^2right)^4=$$
    $$=5sqrt[5]frac(a^2+b^2)^3256a^8b^8geq5sqrt[5]frac(2ab)^3256a^8b^8=5sqrt[5]frac132a^5b^5=frac104abgeqfrac10(a+b)^2.$$






    share|cite|improve this answer

























      up vote
      0
      down vote



      accepted










      Your solution is right.



      The proof of the final inequality we can make also by AM-GM.



      Indeed,
      $$frac1a^2+b^2+frac1a^2+frac1b^2=frac1a^2+b^2+fraca^2+b^2a^2b^2=$$
      $$=frac1a^2+b^2+4cdotfraca^2+b^24a^2b^2geq5sqrt[5]frac1a^2+b^2left(fraca^2+b^24a^2b^2right)^4=$$
      $$=5sqrt[5]frac(a^2+b^2)^3256a^8b^8geq5sqrt[5]frac(2ab)^3256a^8b^8=5sqrt[5]frac132a^5b^5=frac104abgeqfrac10(a+b)^2.$$






      share|cite|improve this answer























        up vote
        0
        down vote



        accepted







        up vote
        0
        down vote



        accepted






        Your solution is right.



        The proof of the final inequality we can make also by AM-GM.



        Indeed,
        $$frac1a^2+b^2+frac1a^2+frac1b^2=frac1a^2+b^2+fraca^2+b^2a^2b^2=$$
        $$=frac1a^2+b^2+4cdotfraca^2+b^24a^2b^2geq5sqrt[5]frac1a^2+b^2left(fraca^2+b^24a^2b^2right)^4=$$
        $$=5sqrt[5]frac(a^2+b^2)^3256a^8b^8geq5sqrt[5]frac(2ab)^3256a^8b^8=5sqrt[5]frac132a^5b^5=frac104abgeqfrac10(a+b)^2.$$






        share|cite|improve this answer













        Your solution is right.



        The proof of the final inequality we can make also by AM-GM.



        Indeed,
        $$frac1a^2+b^2+frac1a^2+frac1b^2=frac1a^2+b^2+fraca^2+b^2a^2b^2=$$
        $$=frac1a^2+b^2+4cdotfraca^2+b^24a^2b^2geq5sqrt[5]frac1a^2+b^2left(fraca^2+b^24a^2b^2right)^4=$$
        $$=5sqrt[5]frac(a^2+b^2)^3256a^8b^8geq5sqrt[5]frac(2ab)^3256a^8b^8=5sqrt[5]frac132a^5b^5=frac104abgeqfrac10(a+b)^2.$$







        share|cite|improve this answer













        share|cite|improve this answer



        share|cite|improve this answer











        answered 2 days ago









        Michael Rozenberg

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