Prove that $frac{1}{2}(x+frac{a}{x}) ge sqrt{a}$












1












$begingroup$


Let x and a be real numbers > 0. Prove that $frac{1}{2}(x+frac{a}{x}) ge sqrt{a}$



My idea is that I'm going to use $a>b iff a^2>b^2$ since we are only dealing with postive real numbers we won't run into problems with the root.



$frac{1}{2}(x+frac{a}{x}) ge sqrt{a} iff (frac{1}{2}(x+frac{a}{x}))^2 = frac{1}{4}(x+frac{a}{x})(x+frac{a}{x})= frac{1}{4}(x^2+2a+frac{a^2}{x^2}) ge sqrt{a}^2=a$



And from that we get:



$frac{1}{4}x^2 +frac{1}{2}a+frac{a^2}{4x^2} ge a$



I don't really know how to proceed from here. What about looking at different possibilities for the values of a and x?



Let x>a. Then:



$frac{1}{4}x^2 +frac{1}{2}a+frac{a^2}{4x^2} ge frac{1}{4}a^2 +frac{1}{2}a+frac{a^2}{4x^2} ge a iff frac{1}{4}a^2x^2 +frac{1}{2}ax^2+frac{a^2}{4} ge ax^2$



And now I don't know how to prcoeed from here and doubt that my approach is correct. Can someone please help me out here? Tanks in advance.






inequality proof-writing







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$endgroup$












  • $begingroup$
    Use AM-GM inequality, like here
    $endgroup$
    – rtybase
    Jan 3 at 23:53












  • $begingroup$
    I don't think the a.m.-g.m.-inequality tag is necessarily appropriate here, as the asker shows no evidence of wanting to use it, or indeed knowing the inequality in the first place. This doesn't appear to be a question about the am-gm inequality.
    $endgroup$
    – Theo Bendit
    Jan 3 at 23:59












  • $begingroup$
    I know the AM-GM inequality but am not allowed to use it
    $endgroup$
    – John D.
    Jan 4 at 0:00










  • $begingroup$
    @JohnD. if you know AM-GM inequality, then you should know how to prove it for 2 numbers. Apply the same strategy
    $endgroup$
    – rtybase
    Jan 4 at 0:05
















1












$begingroup$


Let x and a be real numbers > 0. Prove that $frac{1}{2}(x+frac{a}{x}) ge sqrt{a}$



My idea is that I'm going to use $a>b iff a^2>b^2$ since we are only dealing with postive real numbers we won't run into problems with the root.



$frac{1}{2}(x+frac{a}{x}) ge sqrt{a} iff (frac{1}{2}(x+frac{a}{x}))^2 = frac{1}{4}(x+frac{a}{x})(x+frac{a}{x})= frac{1}{4}(x^2+2a+frac{a^2}{x^2}) ge sqrt{a}^2=a$



And from that we get:



$frac{1}{4}x^2 +frac{1}{2}a+frac{a^2}{4x^2} ge a$



I don't really know how to proceed from here. What about looking at different possibilities for the values of a and x?



Let x>a. Then:



$frac{1}{4}x^2 +frac{1}{2}a+frac{a^2}{4x^2} ge frac{1}{4}a^2 +frac{1}{2}a+frac{a^2}{4x^2} ge a iff frac{1}{4}a^2x^2 +frac{1}{2}ax^2+frac{a^2}{4} ge ax^2$



And now I don't know how to prcoeed from here and doubt that my approach is correct. Can someone please help me out here? Tanks in advance.






inequality proof-writing







share|cite|improve this question











$endgroup$












  • $begingroup$
    Use AM-GM inequality, like here
    $endgroup$
    – rtybase
    Jan 3 at 23:53












  • $begingroup$
    I don't think the a.m.-g.m.-inequality tag is necessarily appropriate here, as the asker shows no evidence of wanting to use it, or indeed knowing the inequality in the first place. This doesn't appear to be a question about the am-gm inequality.
    $endgroup$
    – Theo Bendit
    Jan 3 at 23:59












  • $begingroup$
    I know the AM-GM inequality but am not allowed to use it
    $endgroup$
    – John D.
    Jan 4 at 0:00










  • $begingroup$
    @JohnD. if you know AM-GM inequality, then you should know how to prove it for 2 numbers. Apply the same strategy
    $endgroup$
    – rtybase
    Jan 4 at 0:05














1












1








1





$begingroup$


Let x and a be real numbers > 0. Prove that $frac{1}{2}(x+frac{a}{x}) ge sqrt{a}$



My idea is that I'm going to use $a>b iff a^2>b^2$ since we are only dealing with postive real numbers we won't run into problems with the root.



$frac{1}{2}(x+frac{a}{x}) ge sqrt{a} iff (frac{1}{2}(x+frac{a}{x}))^2 = frac{1}{4}(x+frac{a}{x})(x+frac{a}{x})= frac{1}{4}(x^2+2a+frac{a^2}{x^2}) ge sqrt{a}^2=a$



And from that we get:



$frac{1}{4}x^2 +frac{1}{2}a+frac{a^2}{4x^2} ge a$



I don't really know how to proceed from here. What about looking at different possibilities for the values of a and x?



Let x>a. Then:



$frac{1}{4}x^2 +frac{1}{2}a+frac{a^2}{4x^2} ge frac{1}{4}a^2 +frac{1}{2}a+frac{a^2}{4x^2} ge a iff frac{1}{4}a^2x^2 +frac{1}{2}ax^2+frac{a^2}{4} ge ax^2$



And now I don't know how to prcoeed from here and doubt that my approach is correct. Can someone please help me out here? Tanks in advance.






inequality proof-writing







share|cite|improve this question











$endgroup$




Let x and a be real numbers > 0. Prove that $frac{1}{2}(x+frac{a}{x}) ge sqrt{a}$



My idea is that I'm going to use $a>b iff a^2>b^2$ since we are only dealing with postive real numbers we won't run into problems with the root.



$frac{1}{2}(x+frac{a}{x}) ge sqrt{a} iff (frac{1}{2}(x+frac{a}{x}))^2 = frac{1}{4}(x+frac{a}{x})(x+frac{a}{x})= frac{1}{4}(x^2+2a+frac{a^2}{x^2}) ge sqrt{a}^2=a$



And from that we get:



$frac{1}{4}x^2 +frac{1}{2}a+frac{a^2}{4x^2} ge a$



I don't really know how to proceed from here. What about looking at different possibilities for the values of a and x?



Let x>a. Then:



$frac{1}{4}x^2 +frac{1}{2}a+frac{a^2}{4x^2} ge frac{1}{4}a^2 +frac{1}{2}a+frac{a^2}{4x^2} ge a iff frac{1}{4}a^2x^2 +frac{1}{2}ax^2+frac{a^2}{4} ge ax^2$



And now I don't know how to prcoeed from here and doubt that my approach is correct. Can someone please help me out here? Tanks in advance.






inequality proof-writing




inequality proof-writing






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 3 at 23:59









Theo Bendit

17.2k12149




17.2k12149










asked Jan 3 at 23:50









John D.John D.

153




153












  • $begingroup$
    Use AM-GM inequality, like here
    $endgroup$
    – rtybase
    Jan 3 at 23:53












  • $begingroup$
    I don't think the a.m.-g.m.-inequality tag is necessarily appropriate here, as the asker shows no evidence of wanting to use it, or indeed knowing the inequality in the first place. This doesn't appear to be a question about the am-gm inequality.
    $endgroup$
    – Theo Bendit
    Jan 3 at 23:59












  • $begingroup$
    I know the AM-GM inequality but am not allowed to use it
    $endgroup$
    – John D.
    Jan 4 at 0:00










  • $begingroup$
    @JohnD. if you know AM-GM inequality, then you should know how to prove it for 2 numbers. Apply the same strategy
    $endgroup$
    – rtybase
    Jan 4 at 0:05


















  • $begingroup$
    Use AM-GM inequality, like here
    $endgroup$
    – rtybase
    Jan 3 at 23:53












  • $begingroup$
    I don't think the a.m.-g.m.-inequality tag is necessarily appropriate here, as the asker shows no evidence of wanting to use it, or indeed knowing the inequality in the first place. This doesn't appear to be a question about the am-gm inequality.
    $endgroup$
    – Theo Bendit
    Jan 3 at 23:59












  • $begingroup$
    I know the AM-GM inequality but am not allowed to use it
    $endgroup$
    – John D.
    Jan 4 at 0:00










  • $begingroup$
    @JohnD. if you know AM-GM inequality, then you should know how to prove it for 2 numbers. Apply the same strategy
    $endgroup$
    – rtybase
    Jan 4 at 0:05
















$begingroup$
Use AM-GM inequality, like here
$endgroup$
– rtybase
Jan 3 at 23:53






$begingroup$
Use AM-GM inequality, like here
$endgroup$
– rtybase
Jan 3 at 23:53














$begingroup$
I don't think the a.m.-g.m.-inequality tag is necessarily appropriate here, as the asker shows no evidence of wanting to use it, or indeed knowing the inequality in the first place. This doesn't appear to be a question about the am-gm inequality.
$endgroup$
– Theo Bendit
Jan 3 at 23:59






$begingroup$
I don't think the a.m.-g.m.-inequality tag is necessarily appropriate here, as the asker shows no evidence of wanting to use it, or indeed knowing the inequality in the first place. This doesn't appear to be a question about the am-gm inequality.
$endgroup$
– Theo Bendit
Jan 3 at 23:59














$begingroup$
I know the AM-GM inequality but am not allowed to use it
$endgroup$
– John D.
Jan 4 at 0:00




$begingroup$
I know the AM-GM inequality but am not allowed to use it
$endgroup$
– John D.
Jan 4 at 0:00












$begingroup$
@JohnD. if you know AM-GM inequality, then you should know how to prove it for 2 numbers. Apply the same strategy
$endgroup$
– rtybase
Jan 4 at 0:05




$begingroup$
@JohnD. if you know AM-GM inequality, then you should know how to prove it for 2 numbers. Apply the same strategy
$endgroup$
– rtybase
Jan 4 at 0:05










3 Answers
3






active

oldest

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1












$begingroup$

From here:
$$frac{1}{4}x^2 +frac{1}{2}a+frac{a^2}{4x^2} ge a,$$
move the $a$ to the left side to yield
$$frac{1}{4}x^2 - frac{1}{2}a+frac{a^2}{4x^2} ge 0.$$
The left side is a perfect square:
$$left(frac{x}{2} - frac{a}{2x}right)^2.$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you. I was really hoping someone else saw the obvious, straightforward answer. :) +$1$
    $endgroup$
    – Clayton
    Jan 4 at 0:40



















3












$begingroup$

Because by AM-GM $$x+frac{a}{x}geq2sqrt{xcdotfrac{a}{x}}=2sqrt{a}.$$






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    $(sqrt{x}-sqrt{a/x})^2geq0Rightarrow x+a/x-2sqrt{a}geq 0Rightarrow 1/2(x+a/x)geq sqrt{a}.$






    share|cite|improve this answer









    $endgroup$













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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      1












      $begingroup$

      From here:
      $$frac{1}{4}x^2 +frac{1}{2}a+frac{a^2}{4x^2} ge a,$$
      move the $a$ to the left side to yield
      $$frac{1}{4}x^2 - frac{1}{2}a+frac{a^2}{4x^2} ge 0.$$
      The left side is a perfect square:
      $$left(frac{x}{2} - frac{a}{2x}right)^2.$$






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        Thank you. I was really hoping someone else saw the obvious, straightforward answer. :) +$1$
        $endgroup$
        – Clayton
        Jan 4 at 0:40
















      1












      $begingroup$

      From here:
      $$frac{1}{4}x^2 +frac{1}{2}a+frac{a^2}{4x^2} ge a,$$
      move the $a$ to the left side to yield
      $$frac{1}{4}x^2 - frac{1}{2}a+frac{a^2}{4x^2} ge 0.$$
      The left side is a perfect square:
      $$left(frac{x}{2} - frac{a}{2x}right)^2.$$






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        Thank you. I was really hoping someone else saw the obvious, straightforward answer. :) +$1$
        $endgroup$
        – Clayton
        Jan 4 at 0:40














      1












      1








      1





      $begingroup$

      From here:
      $$frac{1}{4}x^2 +frac{1}{2}a+frac{a^2}{4x^2} ge a,$$
      move the $a$ to the left side to yield
      $$frac{1}{4}x^2 - frac{1}{2}a+frac{a^2}{4x^2} ge 0.$$
      The left side is a perfect square:
      $$left(frac{x}{2} - frac{a}{2x}right)^2.$$






      share|cite|improve this answer









      $endgroup$



      From here:
      $$frac{1}{4}x^2 +frac{1}{2}a+frac{a^2}{4x^2} ge a,$$
      move the $a$ to the left side to yield
      $$frac{1}{4}x^2 - frac{1}{2}a+frac{a^2}{4x^2} ge 0.$$
      The left side is a perfect square:
      $$left(frac{x}{2} - frac{a}{2x}right)^2.$$







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered Jan 3 at 23:53









      Theo BenditTheo Bendit

      17.2k12149




      17.2k12149












      • $begingroup$
        Thank you. I was really hoping someone else saw the obvious, straightforward answer. :) +$1$
        $endgroup$
        – Clayton
        Jan 4 at 0:40


















      • $begingroup$
        Thank you. I was really hoping someone else saw the obvious, straightforward answer. :) +$1$
        $endgroup$
        – Clayton
        Jan 4 at 0:40
















      $begingroup$
      Thank you. I was really hoping someone else saw the obvious, straightforward answer. :) +$1$
      $endgroup$
      – Clayton
      Jan 4 at 0:40




      $begingroup$
      Thank you. I was really hoping someone else saw the obvious, straightforward answer. :) +$1$
      $endgroup$
      – Clayton
      Jan 4 at 0:40











      3












      $begingroup$

      Because by AM-GM $$x+frac{a}{x}geq2sqrt{xcdotfrac{a}{x}}=2sqrt{a}.$$






      share|cite|improve this answer









      $endgroup$


















        3












        $begingroup$

        Because by AM-GM $$x+frac{a}{x}geq2sqrt{xcdotfrac{a}{x}}=2sqrt{a}.$$






        share|cite|improve this answer









        $endgroup$
















          3












          3








          3





          $begingroup$

          Because by AM-GM $$x+frac{a}{x}geq2sqrt{xcdotfrac{a}{x}}=2sqrt{a}.$$






          share|cite|improve this answer









          $endgroup$



          Because by AM-GM $$x+frac{a}{x}geq2sqrt{xcdotfrac{a}{x}}=2sqrt{a}.$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 3 at 23:54









          Michael RozenbergMichael Rozenberg

          98.5k1590189




          98.5k1590189























              1












              $begingroup$

              $(sqrt{x}-sqrt{a/x})^2geq0Rightarrow x+a/x-2sqrt{a}geq 0Rightarrow 1/2(x+a/x)geq sqrt{a}.$






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                $(sqrt{x}-sqrt{a/x})^2geq0Rightarrow x+a/x-2sqrt{a}geq 0Rightarrow 1/2(x+a/x)geq sqrt{a}.$






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  $(sqrt{x}-sqrt{a/x})^2geq0Rightarrow x+a/x-2sqrt{a}geq 0Rightarrow 1/2(x+a/x)geq sqrt{a}.$






                  share|cite|improve this answer









                  $endgroup$



                  $(sqrt{x}-sqrt{a/x})^2geq0Rightarrow x+a/x-2sqrt{a}geq 0Rightarrow 1/2(x+a/x)geq sqrt{a}.$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 3 at 23:55









                  John_WickJohn_Wick

                  1,486111




                  1,486111






























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