Prove that $2sqrt{frac{a}{a+c}}-frac{9sqrt{c^2+1}}{8c}0$












0












$begingroup$


Let $a;c>0$ such that $a^2+2ac=1$. Prove that $$2sqrt{frac{a}{a+c}}-frac{9sqrt{c^2+1}}{8c}<1$$




My proof:



$Leftrightarrow 2sqrt{frac{a}{a+c}}-frac{9sqrt{c^2+a^2+2ac}}{8c}<1$



$Leftrightarrow 2sqrt{frac{a}{a+c}}<1+frac{9left(a+cright)}{8c}$



$Leftrightarrow frac{4a}{a+c}<frac{81a^2+306ac+289c^2}{64c^2}$



$Leftrightarrow frac{81a^3+387a^2c+339ac^2+289c^3}{64c^2left(a+cright)}>0$



The last inequality is true by $a;c>0$. I want some other way and a inequality is rigider than original inequality. Tks






inequality







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    0












    $begingroup$


    Let $a;c>0$ such that $a^2+2ac=1$. Prove that $$2sqrt{frac{a}{a+c}}-frac{9sqrt{c^2+1}}{8c}<1$$




    My proof:



    $Leftrightarrow 2sqrt{frac{a}{a+c}}-frac{9sqrt{c^2+a^2+2ac}}{8c}<1$



    $Leftrightarrow 2sqrt{frac{a}{a+c}}<1+frac{9left(a+cright)}{8c}$



    $Leftrightarrow frac{4a}{a+c}<frac{81a^2+306ac+289c^2}{64c^2}$



    $Leftrightarrow frac{81a^3+387a^2c+339ac^2+289c^3}{64c^2left(a+cright)}>0$



    The last inequality is true by $a;c>0$. I want some other way and a inequality is rigider than original inequality. Tks






    inequality







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      Let $a;c>0$ such that $a^2+2ac=1$. Prove that $$2sqrt{frac{a}{a+c}}-frac{9sqrt{c^2+1}}{8c}<1$$




      My proof:



      $Leftrightarrow 2sqrt{frac{a}{a+c}}-frac{9sqrt{c^2+a^2+2ac}}{8c}<1$



      $Leftrightarrow 2sqrt{frac{a}{a+c}}<1+frac{9left(a+cright)}{8c}$



      $Leftrightarrow frac{4a}{a+c}<frac{81a^2+306ac+289c^2}{64c^2}$



      $Leftrightarrow frac{81a^3+387a^2c+339ac^2+289c^3}{64c^2left(a+cright)}>0$



      The last inequality is true by $a;c>0$. I want some other way and a inequality is rigider than original inequality. Tks






      inequality







      share|cite|improve this question









      $endgroup$




      Let $a;c>0$ such that $a^2+2ac=1$. Prove that $$2sqrt{frac{a}{a+c}}-frac{9sqrt{c^2+1}}{8c}<1$$




      My proof:



      $Leftrightarrow 2sqrt{frac{a}{a+c}}-frac{9sqrt{c^2+a^2+2ac}}{8c}<1$



      $Leftrightarrow 2sqrt{frac{a}{a+c}}<1+frac{9left(a+cright)}{8c}$



      $Leftrightarrow frac{4a}{a+c}<frac{81a^2+306ac+289c^2}{64c^2}$



      $Leftrightarrow frac{81a^3+387a^2c+339ac^2+289c^3}{64c^2left(a+cright)}>0$



      The last inequality is true by $a;c>0$. I want some other way and a inequality is rigider than original inequality. Tks






      inequality




      inequality






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 2 '18 at 15:22









      Nguyễn Duy LinhNguyễn Duy Linh

      1748




      1748






















          1 Answer
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          $begingroup$

          $2sqrt{frac{a}{a+c}}-frac{9sqrt{c^2+1}}{8c}=2sqrt{frac{a+c-c}{a+c}}-frac{9sqrt{a^2+c^2+2ac}}{8c}=2sqrt{1-frac c{a+c}}-frac{9(a+c)}{8c}<2-frac{9c}{8c}=2-frac98=frac78<1$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Do you have some suggest about a rigid inequality like maximize ?
            $endgroup$
            – Nguyễn Duy Linh
            Dec 2 '18 at 23:35










          • $begingroup$
            I used the inequalities $sqrt{1-frac c{a+c}}<1, a+c>c$ which are natural implications of $a,c>0$. What do you mean by rigid inequality? Does it mean an inequality that is true in general?
            $endgroup$
            – Shubham Johri
            Dec 3 '18 at 8:09










          • $begingroup$
            It means i want to know the maximize of this inequality.
            $endgroup$
            – Nguyễn Duy Linh
            Dec 3 '18 at 10:21











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          $begingroup$

          $2sqrt{frac{a}{a+c}}-frac{9sqrt{c^2+1}}{8c}=2sqrt{frac{a+c-c}{a+c}}-frac{9sqrt{a^2+c^2+2ac}}{8c}=2sqrt{1-frac c{a+c}}-frac{9(a+c)}{8c}<2-frac{9c}{8c}=2-frac98=frac78<1$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Do you have some suggest about a rigid inequality like maximize ?
            $endgroup$
            – Nguyễn Duy Linh
            Dec 2 '18 at 23:35










          • $begingroup$
            I used the inequalities $sqrt{1-frac c{a+c}}<1, a+c>c$ which are natural implications of $a,c>0$. What do you mean by rigid inequality? Does it mean an inequality that is true in general?
            $endgroup$
            – Shubham Johri
            Dec 3 '18 at 8:09










          • $begingroup$
            It means i want to know the maximize of this inequality.
            $endgroup$
            – Nguyễn Duy Linh
            Dec 3 '18 at 10:21
















          1












          $begingroup$

          $2sqrt{frac{a}{a+c}}-frac{9sqrt{c^2+1}}{8c}=2sqrt{frac{a+c-c}{a+c}}-frac{9sqrt{a^2+c^2+2ac}}{8c}=2sqrt{1-frac c{a+c}}-frac{9(a+c)}{8c}<2-frac{9c}{8c}=2-frac98=frac78<1$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Do you have some suggest about a rigid inequality like maximize ?
            $endgroup$
            – Nguyễn Duy Linh
            Dec 2 '18 at 23:35










          • $begingroup$
            I used the inequalities $sqrt{1-frac c{a+c}}<1, a+c>c$ which are natural implications of $a,c>0$. What do you mean by rigid inequality? Does it mean an inequality that is true in general?
            $endgroup$
            – Shubham Johri
            Dec 3 '18 at 8:09










          • $begingroup$
            It means i want to know the maximize of this inequality.
            $endgroup$
            – Nguyễn Duy Linh
            Dec 3 '18 at 10:21














          1












          1








          1





          $begingroup$

          $2sqrt{frac{a}{a+c}}-frac{9sqrt{c^2+1}}{8c}=2sqrt{frac{a+c-c}{a+c}}-frac{9sqrt{a^2+c^2+2ac}}{8c}=2sqrt{1-frac c{a+c}}-frac{9(a+c)}{8c}<2-frac{9c}{8c}=2-frac98=frac78<1$






          share|cite|improve this answer









          $endgroup$



          $2sqrt{frac{a}{a+c}}-frac{9sqrt{c^2+1}}{8c}=2sqrt{frac{a+c-c}{a+c}}-frac{9sqrt{a^2+c^2+2ac}}{8c}=2sqrt{1-frac c{a+c}}-frac{9(a+c)}{8c}<2-frac{9c}{8c}=2-frac98=frac78<1$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 2 '18 at 15:40









          Shubham JohriShubham Johri

          4,759717




          4,759717












          • $begingroup$
            Do you have some suggest about a rigid inequality like maximize ?
            $endgroup$
            – Nguyễn Duy Linh
            Dec 2 '18 at 23:35










          • $begingroup$
            I used the inequalities $sqrt{1-frac c{a+c}}<1, a+c>c$ which are natural implications of $a,c>0$. What do you mean by rigid inequality? Does it mean an inequality that is true in general?
            $endgroup$
            – Shubham Johri
            Dec 3 '18 at 8:09










          • $begingroup$
            It means i want to know the maximize of this inequality.
            $endgroup$
            – Nguyễn Duy Linh
            Dec 3 '18 at 10:21


















          • $begingroup$
            Do you have some suggest about a rigid inequality like maximize ?
            $endgroup$
            – Nguyễn Duy Linh
            Dec 2 '18 at 23:35










          • $begingroup$
            I used the inequalities $sqrt{1-frac c{a+c}}<1, a+c>c$ which are natural implications of $a,c>0$. What do you mean by rigid inequality? Does it mean an inequality that is true in general?
            $endgroup$
            – Shubham Johri
            Dec 3 '18 at 8:09










          • $begingroup$
            It means i want to know the maximize of this inequality.
            $endgroup$
            – Nguyễn Duy Linh
            Dec 3 '18 at 10:21
















          $begingroup$
          Do you have some suggest about a rigid inequality like maximize ?
          $endgroup$
          – Nguyễn Duy Linh
          Dec 2 '18 at 23:35




          $begingroup$
          Do you have some suggest about a rigid inequality like maximize ?
          $endgroup$
          – Nguyễn Duy Linh
          Dec 2 '18 at 23:35












          $begingroup$
          I used the inequalities $sqrt{1-frac c{a+c}}<1, a+c>c$ which are natural implications of $a,c>0$. What do you mean by rigid inequality? Does it mean an inequality that is true in general?
          $endgroup$
          – Shubham Johri
          Dec 3 '18 at 8:09




          $begingroup$
          I used the inequalities $sqrt{1-frac c{a+c}}<1, a+c>c$ which are natural implications of $a,c>0$. What do you mean by rigid inequality? Does it mean an inequality that is true in general?
          $endgroup$
          – Shubham Johri
          Dec 3 '18 at 8:09












          $begingroup$
          It means i want to know the maximize of this inequality.
          $endgroup$
          – Nguyễn Duy Linh
          Dec 3 '18 at 10:21




          $begingroup$
          It means i want to know the maximize of this inequality.
          $endgroup$
          – Nguyễn Duy Linh
          Dec 3 '18 at 10:21


















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