Let $a>b$ and $ab=1$ show that $frac{a^2+b^2}{a-b} geq 2sqrt{2}$











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Let $a>b$ and $ab=1$. Show that $$frac{a^2+b^2}{a-b} geq 2sqrt{2}$$



My attempt:
$a-b>0$
$$a^2+b^2geq 2sqrt{2}(a-b)$$
$$frac{a^2+b^2}{2}geq sqrt{2}(a-b)$$
By AG inequality we know that $$frac{a^2+b^2}{2}geq sqrt{a^2b^2}=ab=1$$
But this isn't very helpful.






inequality proof-writing







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    up vote
    1
    down vote

    favorite












    Let $a>b$ and $ab=1$. Show that $$frac{a^2+b^2}{a-b} geq 2sqrt{2}$$



    My attempt:
    $a-b>0$
    $$a^2+b^2geq 2sqrt{2}(a-b)$$
    $$frac{a^2+b^2}{2}geq sqrt{2}(a-b)$$
    By AG inequality we know that $$frac{a^2+b^2}{2}geq sqrt{a^2b^2}=ab=1$$
    But this isn't very helpful.






    inequality proof-writing







    share|cite|improve this question


























      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      Let $a>b$ and $ab=1$. Show that $$frac{a^2+b^2}{a-b} geq 2sqrt{2}$$



      My attempt:
      $a-b>0$
      $$a^2+b^2geq 2sqrt{2}(a-b)$$
      $$frac{a^2+b^2}{2}geq sqrt{2}(a-b)$$
      By AG inequality we know that $$frac{a^2+b^2}{2}geq sqrt{a^2b^2}=ab=1$$
      But this isn't very helpful.






      inequality proof-writing







      share|cite|improve this question















      Let $a>b$ and $ab=1$. Show that $$frac{a^2+b^2}{a-b} geq 2sqrt{2}$$



      My attempt:
      $a-b>0$
      $$a^2+b^2geq 2sqrt{2}(a-b)$$
      $$frac{a^2+b^2}{2}geq sqrt{2}(a-b)$$
      By AG inequality we know that $$frac{a^2+b^2}{2}geq sqrt{a^2b^2}=ab=1$$
      But this isn't very helpful.






      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 11 hours ago









      greedoid

      34.3k114488




      34.3k114488










      asked 12 hours ago









      Lovro Sindičić

      244216




      244216






















          4 Answers
          4






          active

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          up vote
          4
          down vote



          accepted










          You may proceed as follows using AM-GM:



          $$frac{a^2+b^2}{a-b} = frac{a^2+b^2 - 2ab + 2}{a-b} = frac{(a-b)^2 + 2}{a-b}= (a-b)+ frac{2}{a-b} stackrel{AM-GM}{geq}2sqrt{2}$$






          share|cite|improve this answer




























            up vote
            1
            down vote













            Consider the map$$begin{array}{rccc}fcolon&(1,infty)&longrightarrow&mathbb R\&a&mapsto&dfrac{a^2+frac1{a^2}}{a-frac1a}.end{array}$$Then$$f'(a)=frac{left(a^2+1right) left(a^4-4 a^2+1right)}{a^2left(a^2-1right)^2}.$$You can easily deduce from this that the minimum of $f$ is attained when $a=sqrt{2+sqrt3}$. But $fleft(sqrt{2+sqrt3}right)=2sqrt2$.






            share|cite|improve this answer




























              up vote
              0
              down vote













              Without applying any of the inequality identities, you can do it the tedious way using calculus from $b=1/a$ to give $$frac{a^2+b^2}{a-b}=frac{a^2+1/a^2}{a-1/a}=frac{a^4+1}{a(a^2-1)}$$ The stationary point of this occurs when the derivative is zero; that is, $$frac{4a^4(a^2-1)-(a^4+1)(3a^2-1)}{a^2(a^2-1)^2}=0$$ so $$4a^6-4a^4-3a^6+a^4-3a^2+1=a^6-3a^4-3a^2+1=0.$$ Note that $ane0,pm1$ as $a>b$. The roots of the above are equivalent to the square roots of the roots of the corresponding cubic; that is, $$a^2=-1,2-sqrt3,2+sqrt3$$ so $a=sqrt{2+sqrt3}$ and $b=sqrt{2-sqrt3}$ as $a>b$. It is possible to find the second derivative to show that it is a minimum. Hence $$frac{a^2+b^2}{a-b}gefrac{2+sqrt3+2-sqrt3}{sqrt{2+sqrt3}-sqrt{2-sqrt3}}=2sqrt2$$ as required.






              share|cite|improve this answer




























                up vote
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                down vote













                Let $x=a-b>0$, then we have to prove $$(b+x)^2+b^2 geq 2xsqrt{2}$$
                or $$2b^2+2bx+x^2geq 2xsqrt{2}$$
                Since $1=ab = b^2+bx$ we have to prove$$2+x^2geq 2xsqrt{2}$$



                which is the same as $$(x-sqrt{2})^2geq 0$$
                and we are done.






                share|cite|improve this answer























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                  4 Answers
                  4






                  active

                  oldest

                  votes








                  4 Answers
                  4






                  active

                  oldest

                  votes









                  active

                  oldest

                  votes






                  active

                  oldest

                  votes








                  up vote
                  4
                  down vote



                  accepted










                  You may proceed as follows using AM-GM:



                  $$frac{a^2+b^2}{a-b} = frac{a^2+b^2 - 2ab + 2}{a-b} = frac{(a-b)^2 + 2}{a-b}= (a-b)+ frac{2}{a-b} stackrel{AM-GM}{geq}2sqrt{2}$$






                  share|cite|improve this answer

























                    up vote
                    4
                    down vote



                    accepted










                    You may proceed as follows using AM-GM:



                    $$frac{a^2+b^2}{a-b} = frac{a^2+b^2 - 2ab + 2}{a-b} = frac{(a-b)^2 + 2}{a-b}= (a-b)+ frac{2}{a-b} stackrel{AM-GM}{geq}2sqrt{2}$$






                    share|cite|improve this answer























                      up vote
                      4
                      down vote



                      accepted







                      up vote
                      4
                      down vote



                      accepted






                      You may proceed as follows using AM-GM:



                      $$frac{a^2+b^2}{a-b} = frac{a^2+b^2 - 2ab + 2}{a-b} = frac{(a-b)^2 + 2}{a-b}= (a-b)+ frac{2}{a-b} stackrel{AM-GM}{geq}2sqrt{2}$$






                      share|cite|improve this answer












                      You may proceed as follows using AM-GM:



                      $$frac{a^2+b^2}{a-b} = frac{a^2+b^2 - 2ab + 2}{a-b} = frac{(a-b)^2 + 2}{a-b}= (a-b)+ frac{2}{a-b} stackrel{AM-GM}{geq}2sqrt{2}$$







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered 11 hours ago









                      trancelocation

                      8,0561519




                      8,0561519






















                          up vote
                          1
                          down vote













                          Consider the map$$begin{array}{rccc}fcolon&(1,infty)&longrightarrow&mathbb R\&a&mapsto&dfrac{a^2+frac1{a^2}}{a-frac1a}.end{array}$$Then$$f'(a)=frac{left(a^2+1right) left(a^4-4 a^2+1right)}{a^2left(a^2-1right)^2}.$$You can easily deduce from this that the minimum of $f$ is attained when $a=sqrt{2+sqrt3}$. But $fleft(sqrt{2+sqrt3}right)=2sqrt2$.






                          share|cite|improve this answer

























                            up vote
                            1
                            down vote













                            Consider the map$$begin{array}{rccc}fcolon&(1,infty)&longrightarrow&mathbb R\&a&mapsto&dfrac{a^2+frac1{a^2}}{a-frac1a}.end{array}$$Then$$f'(a)=frac{left(a^2+1right) left(a^4-4 a^2+1right)}{a^2left(a^2-1right)^2}.$$You can easily deduce from this that the minimum of $f$ is attained when $a=sqrt{2+sqrt3}$. But $fleft(sqrt{2+sqrt3}right)=2sqrt2$.






                            share|cite|improve this answer























                              up vote
                              1
                              down vote










                              up vote
                              1
                              down vote









                              Consider the map$$begin{array}{rccc}fcolon&(1,infty)&longrightarrow&mathbb R\&a&mapsto&dfrac{a^2+frac1{a^2}}{a-frac1a}.end{array}$$Then$$f'(a)=frac{left(a^2+1right) left(a^4-4 a^2+1right)}{a^2left(a^2-1right)^2}.$$You can easily deduce from this that the minimum of $f$ is attained when $a=sqrt{2+sqrt3}$. But $fleft(sqrt{2+sqrt3}right)=2sqrt2$.






                              share|cite|improve this answer












                              Consider the map$$begin{array}{rccc}fcolon&(1,infty)&longrightarrow&mathbb R\&a&mapsto&dfrac{a^2+frac1{a^2}}{a-frac1a}.end{array}$$Then$$f'(a)=frac{left(a^2+1right) left(a^4-4 a^2+1right)}{a^2left(a^2-1right)^2}.$$You can easily deduce from this that the minimum of $f$ is attained when $a=sqrt{2+sqrt3}$. But $fleft(sqrt{2+sqrt3}right)=2sqrt2$.







                              share|cite|improve this answer












                              share|cite|improve this answer



                              share|cite|improve this answer










                              answered 11 hours ago









                              José Carlos Santos

                              139k18111203




                              139k18111203






















                                  up vote
                                  0
                                  down vote













                                  Without applying any of the inequality identities, you can do it the tedious way using calculus from $b=1/a$ to give $$frac{a^2+b^2}{a-b}=frac{a^2+1/a^2}{a-1/a}=frac{a^4+1}{a(a^2-1)}$$ The stationary point of this occurs when the derivative is zero; that is, $$frac{4a^4(a^2-1)-(a^4+1)(3a^2-1)}{a^2(a^2-1)^2}=0$$ so $$4a^6-4a^4-3a^6+a^4-3a^2+1=a^6-3a^4-3a^2+1=0.$$ Note that $ane0,pm1$ as $a>b$. The roots of the above are equivalent to the square roots of the roots of the corresponding cubic; that is, $$a^2=-1,2-sqrt3,2+sqrt3$$ so $a=sqrt{2+sqrt3}$ and $b=sqrt{2-sqrt3}$ as $a>b$. It is possible to find the second derivative to show that it is a minimum. Hence $$frac{a^2+b^2}{a-b}gefrac{2+sqrt3+2-sqrt3}{sqrt{2+sqrt3}-sqrt{2-sqrt3}}=2sqrt2$$ as required.






                                  share|cite|improve this answer

























                                    up vote
                                    0
                                    down vote













                                    Without applying any of the inequality identities, you can do it the tedious way using calculus from $b=1/a$ to give $$frac{a^2+b^2}{a-b}=frac{a^2+1/a^2}{a-1/a}=frac{a^4+1}{a(a^2-1)}$$ The stationary point of this occurs when the derivative is zero; that is, $$frac{4a^4(a^2-1)-(a^4+1)(3a^2-1)}{a^2(a^2-1)^2}=0$$ so $$4a^6-4a^4-3a^6+a^4-3a^2+1=a^6-3a^4-3a^2+1=0.$$ Note that $ane0,pm1$ as $a>b$. The roots of the above are equivalent to the square roots of the roots of the corresponding cubic; that is, $$a^2=-1,2-sqrt3,2+sqrt3$$ so $a=sqrt{2+sqrt3}$ and $b=sqrt{2-sqrt3}$ as $a>b$. It is possible to find the second derivative to show that it is a minimum. Hence $$frac{a^2+b^2}{a-b}gefrac{2+sqrt3+2-sqrt3}{sqrt{2+sqrt3}-sqrt{2-sqrt3}}=2sqrt2$$ as required.






                                    share|cite|improve this answer























                                      up vote
                                      0
                                      down vote










                                      up vote
                                      0
                                      down vote









                                      Without applying any of the inequality identities, you can do it the tedious way using calculus from $b=1/a$ to give $$frac{a^2+b^2}{a-b}=frac{a^2+1/a^2}{a-1/a}=frac{a^4+1}{a(a^2-1)}$$ The stationary point of this occurs when the derivative is zero; that is, $$frac{4a^4(a^2-1)-(a^4+1)(3a^2-1)}{a^2(a^2-1)^2}=0$$ so $$4a^6-4a^4-3a^6+a^4-3a^2+1=a^6-3a^4-3a^2+1=0.$$ Note that $ane0,pm1$ as $a>b$. The roots of the above are equivalent to the square roots of the roots of the corresponding cubic; that is, $$a^2=-1,2-sqrt3,2+sqrt3$$ so $a=sqrt{2+sqrt3}$ and $b=sqrt{2-sqrt3}$ as $a>b$. It is possible to find the second derivative to show that it is a minimum. Hence $$frac{a^2+b^2}{a-b}gefrac{2+sqrt3+2-sqrt3}{sqrt{2+sqrt3}-sqrt{2-sqrt3}}=2sqrt2$$ as required.






                                      share|cite|improve this answer












                                      Without applying any of the inequality identities, you can do it the tedious way using calculus from $b=1/a$ to give $$frac{a^2+b^2}{a-b}=frac{a^2+1/a^2}{a-1/a}=frac{a^4+1}{a(a^2-1)}$$ The stationary point of this occurs when the derivative is zero; that is, $$frac{4a^4(a^2-1)-(a^4+1)(3a^2-1)}{a^2(a^2-1)^2}=0$$ so $$4a^6-4a^4-3a^6+a^4-3a^2+1=a^6-3a^4-3a^2+1=0.$$ Note that $ane0,pm1$ as $a>b$. The roots of the above are equivalent to the square roots of the roots of the corresponding cubic; that is, $$a^2=-1,2-sqrt3,2+sqrt3$$ so $a=sqrt{2+sqrt3}$ and $b=sqrt{2-sqrt3}$ as $a>b$. It is possible to find the second derivative to show that it is a minimum. Hence $$frac{a^2+b^2}{a-b}gefrac{2+sqrt3+2-sqrt3}{sqrt{2+sqrt3}-sqrt{2-sqrt3}}=2sqrt2$$ as required.







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                                      answered 11 hours ago









                                      TheSimpliFire

                                      11.6k62256




                                      11.6k62256






















                                          up vote
                                          0
                                          down vote













                                          Let $x=a-b>0$, then we have to prove $$(b+x)^2+b^2 geq 2xsqrt{2}$$
                                          or $$2b^2+2bx+x^2geq 2xsqrt{2}$$
                                          Since $1=ab = b^2+bx$ we have to prove$$2+x^2geq 2xsqrt{2}$$



                                          which is the same as $$(x-sqrt{2})^2geq 0$$
                                          and we are done.






                                          share|cite|improve this answer



























                                            up vote
                                            0
                                            down vote













                                            Let $x=a-b>0$, then we have to prove $$(b+x)^2+b^2 geq 2xsqrt{2}$$
                                            or $$2b^2+2bx+x^2geq 2xsqrt{2}$$
                                            Since $1=ab = b^2+bx$ we have to prove$$2+x^2geq 2xsqrt{2}$$



                                            which is the same as $$(x-sqrt{2})^2geq 0$$
                                            and we are done.






                                            share|cite|improve this answer

























                                              up vote
                                              0
                                              down vote










                                              up vote
                                              0
                                              down vote









                                              Let $x=a-b>0$, then we have to prove $$(b+x)^2+b^2 geq 2xsqrt{2}$$
                                              or $$2b^2+2bx+x^2geq 2xsqrt{2}$$
                                              Since $1=ab = b^2+bx$ we have to prove$$2+x^2geq 2xsqrt{2}$$



                                              which is the same as $$(x-sqrt{2})^2geq 0$$
                                              and we are done.






                                              share|cite|improve this answer














                                              Let $x=a-b>0$, then we have to prove $$(b+x)^2+b^2 geq 2xsqrt{2}$$
                                              or $$2b^2+2bx+x^2geq 2xsqrt{2}$$
                                              Since $1=ab = b^2+bx$ we have to prove$$2+x^2geq 2xsqrt{2}$$



                                              which is the same as $$(x-sqrt{2})^2geq 0$$
                                              and we are done.







                                              share|cite|improve this answer














                                              share|cite|improve this answer



                                              share|cite|improve this answer








                                              edited 11 hours ago

























                                              answered 11 hours ago









                                              greedoid

                                              34.3k114488




                                              34.3k114488






























                                                   

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