Is there a neat way to find the sign of a real function with radicals like this?












2












$begingroup$


This simple function:



$$f(x) = sqrt{x+2} - 2sqrt{x+1} + sqrt{x}$$



is negative for $x>0$ (just checked out with a graphic calculator).
But how to "prove" this algebraically?



The function comes out from the problem of finding which of these numbers $sqrt{12} - sqrt{11}$ and $sqrt{11} - sqrt{10}$ is bigger.



[BTW, given that $f(x)$ is negative, it should be that $sqrt{11} - sqrt{10}$ is bigger than $sqrt{12} - sqrt{11}$ !]



Thanks!






real-analysis functions







share|cite|improve this question









$endgroup$












  • $begingroup$
    Since you have the tag "real analysis", do you know about differentiation and integration? Because that would give an elegant answer.
    $endgroup$
    – Mees de Vries
    Jan 8 at 15:10
















2












$begingroup$


This simple function:



$$f(x) = sqrt{x+2} - 2sqrt{x+1} + sqrt{x}$$



is negative for $x>0$ (just checked out with a graphic calculator).
But how to "prove" this algebraically?



The function comes out from the problem of finding which of these numbers $sqrt{12} - sqrt{11}$ and $sqrt{11} - sqrt{10}$ is bigger.



[BTW, given that $f(x)$ is negative, it should be that $sqrt{11} - sqrt{10}$ is bigger than $sqrt{12} - sqrt{11}$ !]



Thanks!






real-analysis functions







share|cite|improve this question









$endgroup$












  • $begingroup$
    Since you have the tag "real analysis", do you know about differentiation and integration? Because that would give an elegant answer.
    $endgroup$
    – Mees de Vries
    Jan 8 at 15:10














2












2








2





$begingroup$


This simple function:



$$f(x) = sqrt{x+2} - 2sqrt{x+1} + sqrt{x}$$



is negative for $x>0$ (just checked out with a graphic calculator).
But how to "prove" this algebraically?



The function comes out from the problem of finding which of these numbers $sqrt{12} - sqrt{11}$ and $sqrt{11} - sqrt{10}$ is bigger.



[BTW, given that $f(x)$ is negative, it should be that $sqrt{11} - sqrt{10}$ is bigger than $sqrt{12} - sqrt{11}$ !]



Thanks!






real-analysis functions







share|cite|improve this question









$endgroup$




This simple function:



$$f(x) = sqrt{x+2} - 2sqrt{x+1} + sqrt{x}$$



is negative for $x>0$ (just checked out with a graphic calculator).
But how to "prove" this algebraically?



The function comes out from the problem of finding which of these numbers $sqrt{12} - sqrt{11}$ and $sqrt{11} - sqrt{10}$ is bigger.



[BTW, given that $f(x)$ is negative, it should be that $sqrt{11} - sqrt{10}$ is bigger than $sqrt{12} - sqrt{11}$ !]



Thanks!






real-analysis functions




real-analysis functions






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 8 at 15:03









RNaniRNani

132




132












  • $begingroup$
    Since you have the tag "real analysis", do you know about differentiation and integration? Because that would give an elegant answer.
    $endgroup$
    – Mees de Vries
    Jan 8 at 15:10


















  • $begingroup$
    Since you have the tag "real analysis", do you know about differentiation and integration? Because that would give an elegant answer.
    $endgroup$
    – Mees de Vries
    Jan 8 at 15:10
















$begingroup$
Since you have the tag "real analysis", do you know about differentiation and integration? Because that would give an elegant answer.
$endgroup$
– Mees de Vries
Jan 8 at 15:10




$begingroup$
Since you have the tag "real analysis", do you know about differentiation and integration? Because that would give an elegant answer.
$endgroup$
– Mees de Vries
Jan 8 at 15:10










5 Answers
5






active

oldest

votes


















4












$begingroup$

begin{align}
f(x) &= sqrt{x+2}-2sqrt{x+1}+sqrt{x}\
&= sqrt{x+2}-sqrt{x+1}-(sqrt{x+1}-sqrt{x})\
&=frac{1}{sqrt{x+2}+sqrt{x+1}}-frac1{sqrt{x+1}+sqrt{x}}\
&<0
end{align}



since $x+2 > x$.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    (+1) You were too fast for me.
    $endgroup$
    – José Carlos Santos
    Jan 8 at 15:15










  • $begingroup$
    Wow! Thanks a lot! Is this a "standard pattern" to solve these kind of problems? I'm not a mathematician :)
    $endgroup$
    – RNani
    Jan 8 at 15:15






  • 1




    $begingroup$
    I think It's natural when we see subtraction of square root, we think of conjugate.
    $endgroup$
    – Siong Thye Goh
    Jan 8 at 15:17



















1












$begingroup$

I will prove $f(x)<0$



So i need to prove $sqrt{x+2}-2sqrt{x+1}+sqrt{x}<0left(x>0right)$



Or $sqrt{x+2}+sqrt{x}<2sqrt{x+1}$



By square both sides we have: $sqrt{x^2+2x}<x+1$



Or $x^2+2x<x^2+2x+1Leftrightarrow0<1$



So $f(x)<0$ it means $f(x)$ is negative for $x>0$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Interesting. Thanks!!!
    $endgroup$
    – RNani
    Jan 8 at 15:36



















1












$begingroup$

With some analysis



The square root function is strictly concave, hence
$$sqrt{tfrac12 x+tfrac12(x+2)}=sqrt{x+1}>tfrac12sqrt x+tfrac12sqrt{x+2}iff sqrt x+sqrt{x+2}<2sqrt{x+1}.$$






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    Here is a calculus-based answer, that I find more intuitive. We want to prove that
    $$
    sqrt{x+1} - sqrt{x} > sqrt{x+2} - sqrt{x + 1}.
    $$



    Fix $x > 0$. Then the fundamental theorem of calculus tells us that what we want to prove is equivalent to
    $$
    int_{x}^{x+1} frac1{2sqrt{y}} mathrm dy > int_{x+1}^{x+2} frac1{2sqrt{y}} mathrm dy.
    $$
    But because the function $sqrt{y}$ is strictly increasing, the function $frac1{2sqrt{y}}$ is strictly decreasing. In particular, integrating over an interval of length one gives lower numbers as the lower bound of said interval increases.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thanks for your effort. If I had to explain the solution of the problem to an high school student, I would not be sure to go your way :). But anyway... quite elegant indeed.
      $endgroup$
      – RNani
      Jan 8 at 15:27



















    0












    $begingroup$

    $(x+2)^{1/2}-(x+1)^{1/2} - ((x+1)^{1/2} -x^{1/2})$.



    MVT:



    $(x+k+1)^{1/2}-(x+k)^{1/2}=$



    $(1/2)(frac{1}{a})^{1/2}×1,$ where



    $a in (x+k, x+k+1)$.



    Compare the terms for $k=0,1$



    Used:



    $dfrac {f(x)-f(x_o)}{x-x_o}=f'(a)$, where



    $a in (min(x_0,x),max(x_0,x))$.






    share|cite|improve this answer











    $endgroup$













      Your Answer





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      5 Answers
      5






      active

      oldest

      votes








      5 Answers
      5






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      4












      $begingroup$

      begin{align}
      f(x) &= sqrt{x+2}-2sqrt{x+1}+sqrt{x}\
      &= sqrt{x+2}-sqrt{x+1}-(sqrt{x+1}-sqrt{x})\
      &=frac{1}{sqrt{x+2}+sqrt{x+1}}-frac1{sqrt{x+1}+sqrt{x}}\
      &<0
      end{align}



      since $x+2 > x$.






      share|cite|improve this answer









      $endgroup$









      • 1




        $begingroup$
        (+1) You were too fast for me.
        $endgroup$
        – José Carlos Santos
        Jan 8 at 15:15










      • $begingroup$
        Wow! Thanks a lot! Is this a "standard pattern" to solve these kind of problems? I'm not a mathematician :)
        $endgroup$
        – RNani
        Jan 8 at 15:15






      • 1




        $begingroup$
        I think It's natural when we see subtraction of square root, we think of conjugate.
        $endgroup$
        – Siong Thye Goh
        Jan 8 at 15:17
















      4












      $begingroup$

      begin{align}
      f(x) &= sqrt{x+2}-2sqrt{x+1}+sqrt{x}\
      &= sqrt{x+2}-sqrt{x+1}-(sqrt{x+1}-sqrt{x})\
      &=frac{1}{sqrt{x+2}+sqrt{x+1}}-frac1{sqrt{x+1}+sqrt{x}}\
      &<0
      end{align}



      since $x+2 > x$.






      share|cite|improve this answer









      $endgroup$









      • 1




        $begingroup$
        (+1) You were too fast for me.
        $endgroup$
        – José Carlos Santos
        Jan 8 at 15:15










      • $begingroup$
        Wow! Thanks a lot! Is this a "standard pattern" to solve these kind of problems? I'm not a mathematician :)
        $endgroup$
        – RNani
        Jan 8 at 15:15






      • 1




        $begingroup$
        I think It's natural when we see subtraction of square root, we think of conjugate.
        $endgroup$
        – Siong Thye Goh
        Jan 8 at 15:17














      4












      4








      4





      $begingroup$

      begin{align}
      f(x) &= sqrt{x+2}-2sqrt{x+1}+sqrt{x}\
      &= sqrt{x+2}-sqrt{x+1}-(sqrt{x+1}-sqrt{x})\
      &=frac{1}{sqrt{x+2}+sqrt{x+1}}-frac1{sqrt{x+1}+sqrt{x}}\
      &<0
      end{align}



      since $x+2 > x$.






      share|cite|improve this answer









      $endgroup$



      begin{align}
      f(x) &= sqrt{x+2}-2sqrt{x+1}+sqrt{x}\
      &= sqrt{x+2}-sqrt{x+1}-(sqrt{x+1}-sqrt{x})\
      &=frac{1}{sqrt{x+2}+sqrt{x+1}}-frac1{sqrt{x+1}+sqrt{x}}\
      &<0
      end{align}



      since $x+2 > x$.







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered Jan 8 at 15:11









      Siong Thye GohSiong Thye Goh

      100k1466117




      100k1466117








      • 1




        $begingroup$
        (+1) You were too fast for me.
        $endgroup$
        – José Carlos Santos
        Jan 8 at 15:15










      • $begingroup$
        Wow! Thanks a lot! Is this a "standard pattern" to solve these kind of problems? I'm not a mathematician :)
        $endgroup$
        – RNani
        Jan 8 at 15:15






      • 1




        $begingroup$
        I think It's natural when we see subtraction of square root, we think of conjugate.
        $endgroup$
        – Siong Thye Goh
        Jan 8 at 15:17














      • 1




        $begingroup$
        (+1) You were too fast for me.
        $endgroup$
        – José Carlos Santos
        Jan 8 at 15:15










      • $begingroup$
        Wow! Thanks a lot! Is this a "standard pattern" to solve these kind of problems? I'm not a mathematician :)
        $endgroup$
        – RNani
        Jan 8 at 15:15






      • 1




        $begingroup$
        I think It's natural when we see subtraction of square root, we think of conjugate.
        $endgroup$
        – Siong Thye Goh
        Jan 8 at 15:17








      1




      1




      $begingroup$
      (+1) You were too fast for me.
      $endgroup$
      – José Carlos Santos
      Jan 8 at 15:15




      $begingroup$
      (+1) You were too fast for me.
      $endgroup$
      – José Carlos Santos
      Jan 8 at 15:15












      $begingroup$
      Wow! Thanks a lot! Is this a "standard pattern" to solve these kind of problems? I'm not a mathematician :)
      $endgroup$
      – RNani
      Jan 8 at 15:15




      $begingroup$
      Wow! Thanks a lot! Is this a "standard pattern" to solve these kind of problems? I'm not a mathematician :)
      $endgroup$
      – RNani
      Jan 8 at 15:15




      1




      1




      $begingroup$
      I think It's natural when we see subtraction of square root, we think of conjugate.
      $endgroup$
      – Siong Thye Goh
      Jan 8 at 15:17




      $begingroup$
      I think It's natural when we see subtraction of square root, we think of conjugate.
      $endgroup$
      – Siong Thye Goh
      Jan 8 at 15:17











      1












      $begingroup$

      I will prove $f(x)<0$



      So i need to prove $sqrt{x+2}-2sqrt{x+1}+sqrt{x}<0left(x>0right)$



      Or $sqrt{x+2}+sqrt{x}<2sqrt{x+1}$



      By square both sides we have: $sqrt{x^2+2x}<x+1$



      Or $x^2+2x<x^2+2x+1Leftrightarrow0<1$



      So $f(x)<0$ it means $f(x)$ is negative for $x>0$






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        Interesting. Thanks!!!
        $endgroup$
        – RNani
        Jan 8 at 15:36
















      1












      $begingroup$

      I will prove $f(x)<0$



      So i need to prove $sqrt{x+2}-2sqrt{x+1}+sqrt{x}<0left(x>0right)$



      Or $sqrt{x+2}+sqrt{x}<2sqrt{x+1}$



      By square both sides we have: $sqrt{x^2+2x}<x+1$



      Or $x^2+2x<x^2+2x+1Leftrightarrow0<1$



      So $f(x)<0$ it means $f(x)$ is negative for $x>0$






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        Interesting. Thanks!!!
        $endgroup$
        – RNani
        Jan 8 at 15:36














      1












      1








      1





      $begingroup$

      I will prove $f(x)<0$



      So i need to prove $sqrt{x+2}-2sqrt{x+1}+sqrt{x}<0left(x>0right)$



      Or $sqrt{x+2}+sqrt{x}<2sqrt{x+1}$



      By square both sides we have: $sqrt{x^2+2x}<x+1$



      Or $x^2+2x<x^2+2x+1Leftrightarrow0<1$



      So $f(x)<0$ it means $f(x)$ is negative for $x>0$






      share|cite|improve this answer









      $endgroup$



      I will prove $f(x)<0$



      So i need to prove $sqrt{x+2}-2sqrt{x+1}+sqrt{x}<0left(x>0right)$



      Or $sqrt{x+2}+sqrt{x}<2sqrt{x+1}$



      By square both sides we have: $sqrt{x^2+2x}<x+1$



      Or $x^2+2x<x^2+2x+1Leftrightarrow0<1$



      So $f(x)<0$ it means $f(x)$ is negative for $x>0$







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered Jan 8 at 15:17









      Word ShallowWord Shallow

      788518




      788518












      • $begingroup$
        Interesting. Thanks!!!
        $endgroup$
        – RNani
        Jan 8 at 15:36


















      • $begingroup$
        Interesting. Thanks!!!
        $endgroup$
        – RNani
        Jan 8 at 15:36
















      $begingroup$
      Interesting. Thanks!!!
      $endgroup$
      – RNani
      Jan 8 at 15:36




      $begingroup$
      Interesting. Thanks!!!
      $endgroup$
      – RNani
      Jan 8 at 15:36











      1












      $begingroup$

      With some analysis



      The square root function is strictly concave, hence
      $$sqrt{tfrac12 x+tfrac12(x+2)}=sqrt{x+1}>tfrac12sqrt x+tfrac12sqrt{x+2}iff sqrt x+sqrt{x+2}<2sqrt{x+1}.$$






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        With some analysis



        The square root function is strictly concave, hence
        $$sqrt{tfrac12 x+tfrac12(x+2)}=sqrt{x+1}>tfrac12sqrt x+tfrac12sqrt{x+2}iff sqrt x+sqrt{x+2}<2sqrt{x+1}.$$






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          With some analysis



          The square root function is strictly concave, hence
          $$sqrt{tfrac12 x+tfrac12(x+2)}=sqrt{x+1}>tfrac12sqrt x+tfrac12sqrt{x+2}iff sqrt x+sqrt{x+2}<2sqrt{x+1}.$$






          share|cite|improve this answer









          $endgroup$



          With some analysis



          The square root function is strictly concave, hence
          $$sqrt{tfrac12 x+tfrac12(x+2)}=sqrt{x+1}>tfrac12sqrt x+tfrac12sqrt{x+2}iff sqrt x+sqrt{x+2}<2sqrt{x+1}.$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 8 at 15:29









          BernardBernard

          119k740113




          119k740113























              0












              $begingroup$

              Here is a calculus-based answer, that I find more intuitive. We want to prove that
              $$
              sqrt{x+1} - sqrt{x} > sqrt{x+2} - sqrt{x + 1}.
              $$



              Fix $x > 0$. Then the fundamental theorem of calculus tells us that what we want to prove is equivalent to
              $$
              int_{x}^{x+1} frac1{2sqrt{y}} mathrm dy > int_{x+1}^{x+2} frac1{2sqrt{y}} mathrm dy.
              $$
              But because the function $sqrt{y}$ is strictly increasing, the function $frac1{2sqrt{y}}$ is strictly decreasing. In particular, integrating over an interval of length one gives lower numbers as the lower bound of said interval increases.






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                Thanks for your effort. If I had to explain the solution of the problem to an high school student, I would not be sure to go your way :). But anyway... quite elegant indeed.
                $endgroup$
                – RNani
                Jan 8 at 15:27
















              0












              $begingroup$

              Here is a calculus-based answer, that I find more intuitive. We want to prove that
              $$
              sqrt{x+1} - sqrt{x} > sqrt{x+2} - sqrt{x + 1}.
              $$



              Fix $x > 0$. Then the fundamental theorem of calculus tells us that what we want to prove is equivalent to
              $$
              int_{x}^{x+1} frac1{2sqrt{y}} mathrm dy > int_{x+1}^{x+2} frac1{2sqrt{y}} mathrm dy.
              $$
              But because the function $sqrt{y}$ is strictly increasing, the function $frac1{2sqrt{y}}$ is strictly decreasing. In particular, integrating over an interval of length one gives lower numbers as the lower bound of said interval increases.






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                Thanks for your effort. If I had to explain the solution of the problem to an high school student, I would not be sure to go your way :). But anyway... quite elegant indeed.
                $endgroup$
                – RNani
                Jan 8 at 15:27














              0












              0








              0





              $begingroup$

              Here is a calculus-based answer, that I find more intuitive. We want to prove that
              $$
              sqrt{x+1} - sqrt{x} > sqrt{x+2} - sqrt{x + 1}.
              $$



              Fix $x > 0$. Then the fundamental theorem of calculus tells us that what we want to prove is equivalent to
              $$
              int_{x}^{x+1} frac1{2sqrt{y}} mathrm dy > int_{x+1}^{x+2} frac1{2sqrt{y}} mathrm dy.
              $$
              But because the function $sqrt{y}$ is strictly increasing, the function $frac1{2sqrt{y}}$ is strictly decreasing. In particular, integrating over an interval of length one gives lower numbers as the lower bound of said interval increases.






              share|cite|improve this answer









              $endgroup$



              Here is a calculus-based answer, that I find more intuitive. We want to prove that
              $$
              sqrt{x+1} - sqrt{x} > sqrt{x+2} - sqrt{x + 1}.
              $$



              Fix $x > 0$. Then the fundamental theorem of calculus tells us that what we want to prove is equivalent to
              $$
              int_{x}^{x+1} frac1{2sqrt{y}} mathrm dy > int_{x+1}^{x+2} frac1{2sqrt{y}} mathrm dy.
              $$
              But because the function $sqrt{y}$ is strictly increasing, the function $frac1{2sqrt{y}}$ is strictly decreasing. In particular, integrating over an interval of length one gives lower numbers as the lower bound of said interval increases.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Jan 8 at 15:19









              Mees de VriesMees de Vries

              16.6k12654




              16.6k12654












              • $begingroup$
                Thanks for your effort. If I had to explain the solution of the problem to an high school student, I would not be sure to go your way :). But anyway... quite elegant indeed.
                $endgroup$
                – RNani
                Jan 8 at 15:27


















              • $begingroup$
                Thanks for your effort. If I had to explain the solution of the problem to an high school student, I would not be sure to go your way :). But anyway... quite elegant indeed.
                $endgroup$
                – RNani
                Jan 8 at 15:27
















              $begingroup$
              Thanks for your effort. If I had to explain the solution of the problem to an high school student, I would not be sure to go your way :). But anyway... quite elegant indeed.
              $endgroup$
              – RNani
              Jan 8 at 15:27




              $begingroup$
              Thanks for your effort. If I had to explain the solution of the problem to an high school student, I would not be sure to go your way :). But anyway... quite elegant indeed.
              $endgroup$
              – RNani
              Jan 8 at 15:27











              0












              $begingroup$

              $(x+2)^{1/2}-(x+1)^{1/2} - ((x+1)^{1/2} -x^{1/2})$.



              MVT:



              $(x+k+1)^{1/2}-(x+k)^{1/2}=$



              $(1/2)(frac{1}{a})^{1/2}×1,$ where



              $a in (x+k, x+k+1)$.



              Compare the terms for $k=0,1$



              Used:



              $dfrac {f(x)-f(x_o)}{x-x_o}=f'(a)$, where



              $a in (min(x_0,x),max(x_0,x))$.






              share|cite|improve this answer











              $endgroup$


















                0












                $begingroup$

                $(x+2)^{1/2}-(x+1)^{1/2} - ((x+1)^{1/2} -x^{1/2})$.



                MVT:



                $(x+k+1)^{1/2}-(x+k)^{1/2}=$



                $(1/2)(frac{1}{a})^{1/2}×1,$ where



                $a in (x+k, x+k+1)$.



                Compare the terms for $k=0,1$



                Used:



                $dfrac {f(x)-f(x_o)}{x-x_o}=f'(a)$, where



                $a in (min(x_0,x),max(x_0,x))$.






                share|cite|improve this answer











                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  $(x+2)^{1/2}-(x+1)^{1/2} - ((x+1)^{1/2} -x^{1/2})$.



                  MVT:



                  $(x+k+1)^{1/2}-(x+k)^{1/2}=$



                  $(1/2)(frac{1}{a})^{1/2}×1,$ where



                  $a in (x+k, x+k+1)$.



                  Compare the terms for $k=0,1$



                  Used:



                  $dfrac {f(x)-f(x_o)}{x-x_o}=f'(a)$, where



                  $a in (min(x_0,x),max(x_0,x))$.






                  share|cite|improve this answer











                  $endgroup$



                  $(x+2)^{1/2}-(x+1)^{1/2} - ((x+1)^{1/2} -x^{1/2})$.



                  MVT:



                  $(x+k+1)^{1/2}-(x+k)^{1/2}=$



                  $(1/2)(frac{1}{a})^{1/2}×1,$ where



                  $a in (x+k, x+k+1)$.



                  Compare the terms for $k=0,1$



                  Used:



                  $dfrac {f(x)-f(x_o)}{x-x_o}=f'(a)$, where



                  $a in (min(x_0,x),max(x_0,x))$.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Jan 8 at 15:30

























                  answered Jan 8 at 15:24









                  Peter SzilasPeter Szilas

                  11.1k2821




                  11.1k2821






























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