Show that $sqrt{x-1}+sqrt{y-1}leq sqrt{xy}$












2












$begingroup$


How can one show that $sqrt{x-1}+sqrt{y-1}leq sqrt{xy}$



Assuming that :



$sqrt{x-1}+sqrt{y-1}leq sqrt{xy}$



So



$(sqrt{x-1}+sqrt{y-1})^2leq xy$



$sqrt{(x-1)(y-1)} leq xy-x-y+2$



$ (y-1)(x-1)+3 leq sqrt{(x-1)(y-1)}$



Here I'm stuck !






algebra-precalculus inequality radicals







share|cite|improve this question











$endgroup$












  • $begingroup$
    $LaTeX$ hint: If you put braces around everything that belongs under the square root sign, the top bar extends to cover it. Putting sqrt {x-1} between dollar signs gives $sqrt {x-1}$
    $endgroup$
    – Ross Millikan
    Nov 8 '15 at 21:13












  • $begingroup$
    Assuming what needs to be proved is never a good idea. (A valid method of proof is proof by contradiction, which would begin by assuming the opposite of what needs to be proved, and reach a contradiction from there.)
    $endgroup$
    – user147263
    Nov 8 '15 at 21:25


















2












$begingroup$


How can one show that $sqrt{x-1}+sqrt{y-1}leq sqrt{xy}$



Assuming that :



$sqrt{x-1}+sqrt{y-1}leq sqrt{xy}$



So



$(sqrt{x-1}+sqrt{y-1})^2leq xy$



$sqrt{(x-1)(y-1)} leq xy-x-y+2$



$ (y-1)(x-1)+3 leq sqrt{(x-1)(y-1)}$



Here I'm stuck !






algebra-precalculus inequality radicals







share|cite|improve this question











$endgroup$












  • $begingroup$
    $LaTeX$ hint: If you put braces around everything that belongs under the square root sign, the top bar extends to cover it. Putting sqrt {x-1} between dollar signs gives $sqrt {x-1}$
    $endgroup$
    – Ross Millikan
    Nov 8 '15 at 21:13












  • $begingroup$
    Assuming what needs to be proved is never a good idea. (A valid method of proof is proof by contradiction, which would begin by assuming the opposite of what needs to be proved, and reach a contradiction from there.)
    $endgroup$
    – user147263
    Nov 8 '15 at 21:25
















2












2








2





$begingroup$


How can one show that $sqrt{x-1}+sqrt{y-1}leq sqrt{xy}$



Assuming that :



$sqrt{x-1}+sqrt{y-1}leq sqrt{xy}$



So



$(sqrt{x-1}+sqrt{y-1})^2leq xy$



$sqrt{(x-1)(y-1)} leq xy-x-y+2$



$ (y-1)(x-1)+3 leq sqrt{(x-1)(y-1)}$



Here I'm stuck !






algebra-precalculus inequality radicals







share|cite|improve this question











$endgroup$




How can one show that $sqrt{x-1}+sqrt{y-1}leq sqrt{xy}$



Assuming that :



$sqrt{x-1}+sqrt{y-1}leq sqrt{xy}$



So



$(sqrt{x-1}+sqrt{y-1})^2leq xy$



$sqrt{(x-1)(y-1)} leq xy-x-y+2$



$ (y-1)(x-1)+3 leq sqrt{(x-1)(y-1)}$



Here I'm stuck !






algebra-precalculus inequality radicals




algebra-precalculus inequality radicals






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 8 '15 at 21:25







user147263

















asked Nov 8 '15 at 21:10









user233658user233658

66649




66649












  • $begingroup$
    $LaTeX$ hint: If you put braces around everything that belongs under the square root sign, the top bar extends to cover it. Putting sqrt {x-1} between dollar signs gives $sqrt {x-1}$
    $endgroup$
    – Ross Millikan
    Nov 8 '15 at 21:13












  • $begingroup$
    Assuming what needs to be proved is never a good idea. (A valid method of proof is proof by contradiction, which would begin by assuming the opposite of what needs to be proved, and reach a contradiction from there.)
    $endgroup$
    – user147263
    Nov 8 '15 at 21:25




















  • $begingroup$
    $LaTeX$ hint: If you put braces around everything that belongs under the square root sign, the top bar extends to cover it. Putting sqrt {x-1} between dollar signs gives $sqrt {x-1}$
    $endgroup$
    – Ross Millikan
    Nov 8 '15 at 21:13












  • $begingroup$
    Assuming what needs to be proved is never a good idea. (A valid method of proof is proof by contradiction, which would begin by assuming the opposite of what needs to be proved, and reach a contradiction from there.)
    $endgroup$
    – user147263
    Nov 8 '15 at 21:25


















$begingroup$
$LaTeX$ hint: If you put braces around everything that belongs under the square root sign, the top bar extends to cover it. Putting sqrt {x-1} between dollar signs gives $sqrt {x-1}$
$endgroup$
– Ross Millikan
Nov 8 '15 at 21:13






$begingroup$
$LaTeX$ hint: If you put braces around everything that belongs under the square root sign, the top bar extends to cover it. Putting sqrt {x-1} between dollar signs gives $sqrt {x-1}$
$endgroup$
– Ross Millikan
Nov 8 '15 at 21:13














$begingroup$
Assuming what needs to be proved is never a good idea. (A valid method of proof is proof by contradiction, which would begin by assuming the opposite of what needs to be proved, and reach a contradiction from there.)
$endgroup$
– user147263
Nov 8 '15 at 21:25






$begingroup$
Assuming what needs to be proved is never a good idea. (A valid method of proof is proof by contradiction, which would begin by assuming the opposite of what needs to be proved, and reach a contradiction from there.)
$endgroup$
– user147263
Nov 8 '15 at 21:25












3 Answers
3






active

oldest

votes


















3












$begingroup$

There are some errors in your calculation, e.g. a missing factor 2 in
$$
(sqrt{x-1}+sqrt{y-1})^2 = x - 1 + y - 1 + 2sqrt{x-1}sqrt{y-1}
$$
and in the last step the inequality sign is in the wrong direction and
the number $3$ is wrong.




For $x ge 1$, $y ge 1$ you can square the inequality (since both
sides are non-negative):
$$
sqrt{mathstrut x-1}+sqrt{mathstrut y-1}leq sqrt{mathstrut xy} \
Longleftrightarrow (x-1) + (y-1) + 2 sqrt{mathstrut x-1}sqrt{mathstrut y-1} le xy \
Longleftrightarrow 0 le xy - x - y + 2 - 2 sqrt{mathstrut (x-1)(y-1)} \
Longleftrightarrow 0 le (x-1)(y-1) - 2 sqrt{mathstrut (x-1)(y-1)} + 1
$$
With $t := sqrt{(x-1)(y-1)}$ the right-hand side is
$$
t^2 - 2 t + 1 = (t-1)^2 ge 0 , .
$$
so that the inequality is true.



It follows also that equality holds if and only if $t = 1$,
i.e. if $(x-1)(y-1) = 1$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you so much for the answer ! :)
    $endgroup$
    – user233658
    Nov 8 '15 at 21:38










  • $begingroup$
    @user233658: You are welcome!
    $endgroup$
    – Martin R
    Nov 8 '15 at 21:43



















1












$begingroup$

Starting with “Assuming that $X$” when $X$ is the thing to be proved is not the right thing to do. You can make your computations easier if you set
$$
t=sqrt{x-1},quad u=sqrt{y-1}
$$
so $x=t^2+1$ and $y=u^2+1$. The inequality to be proved becomes
$$
t+ulesqrt{(t^2+1)(u^2+1)}
$$
that has a single radical. Since $t+uge0$, the inequality is equivalent to
$$
(t+u)^2le(t^2+1)(u^2+1)
$$
that is,
$$
t^2+2tu+u^2le t^2u^2+t^2+u^2+1
$$
and so equivalent to
$$
0le t^2u^2-2tu+1
$$
(by transporting terms to the right-hand side) which becomes
$$
0le(tu-1)^2
$$
which is true.



Don't forget the $2$ when you square! You basically did $(a+b)^2=a^2+ab+b^2$ and this is where you got stuck.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    WLOG $sqrt x=sec A,sqrt y=sec B$ where $0<A,B<dfracpi2$



    $sqrt{x-1}+sqrt{y-1}=tan A+tan B=dfrac{sin(A+B)}{cos Acos B}lesqrt{sec Asec B}=?$






    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









      3












      $begingroup$

      There are some errors in your calculation, e.g. a missing factor 2 in
      $$
      (sqrt{x-1}+sqrt{y-1})^2 = x - 1 + y - 1 + 2sqrt{x-1}sqrt{y-1}
      $$
      and in the last step the inequality sign is in the wrong direction and
      the number $3$ is wrong.




      For $x ge 1$, $y ge 1$ you can square the inequality (since both
      sides are non-negative):
      $$
      sqrt{mathstrut x-1}+sqrt{mathstrut y-1}leq sqrt{mathstrut xy} \
      Longleftrightarrow (x-1) + (y-1) + 2 sqrt{mathstrut x-1}sqrt{mathstrut y-1} le xy \
      Longleftrightarrow 0 le xy - x - y + 2 - 2 sqrt{mathstrut (x-1)(y-1)} \
      Longleftrightarrow 0 le (x-1)(y-1) - 2 sqrt{mathstrut (x-1)(y-1)} + 1
      $$
      With $t := sqrt{(x-1)(y-1)}$ the right-hand side is
      $$
      t^2 - 2 t + 1 = (t-1)^2 ge 0 , .
      $$
      so that the inequality is true.



      It follows also that equality holds if and only if $t = 1$,
      i.e. if $(x-1)(y-1) = 1$.






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        Thank you so much for the answer ! :)
        $endgroup$
        – user233658
        Nov 8 '15 at 21:38










      • $begingroup$
        @user233658: You are welcome!
        $endgroup$
        – Martin R
        Nov 8 '15 at 21:43
















      3












      $begingroup$

      There are some errors in your calculation, e.g. a missing factor 2 in
      $$
      (sqrt{x-1}+sqrt{y-1})^2 = x - 1 + y - 1 + 2sqrt{x-1}sqrt{y-1}
      $$
      and in the last step the inequality sign is in the wrong direction and
      the number $3$ is wrong.




      For $x ge 1$, $y ge 1$ you can square the inequality (since both
      sides are non-negative):
      $$
      sqrt{mathstrut x-1}+sqrt{mathstrut y-1}leq sqrt{mathstrut xy} \
      Longleftrightarrow (x-1) + (y-1) + 2 sqrt{mathstrut x-1}sqrt{mathstrut y-1} le xy \
      Longleftrightarrow 0 le xy - x - y + 2 - 2 sqrt{mathstrut (x-1)(y-1)} \
      Longleftrightarrow 0 le (x-1)(y-1) - 2 sqrt{mathstrut (x-1)(y-1)} + 1
      $$
      With $t := sqrt{(x-1)(y-1)}$ the right-hand side is
      $$
      t^2 - 2 t + 1 = (t-1)^2 ge 0 , .
      $$
      so that the inequality is true.



      It follows also that equality holds if and only if $t = 1$,
      i.e. if $(x-1)(y-1) = 1$.






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        Thank you so much for the answer ! :)
        $endgroup$
        – user233658
        Nov 8 '15 at 21:38










      • $begingroup$
        @user233658: You are welcome!
        $endgroup$
        – Martin R
        Nov 8 '15 at 21:43














      3












      3








      3





      $begingroup$

      There are some errors in your calculation, e.g. a missing factor 2 in
      $$
      (sqrt{x-1}+sqrt{y-1})^2 = x - 1 + y - 1 + 2sqrt{x-1}sqrt{y-1}
      $$
      and in the last step the inequality sign is in the wrong direction and
      the number $3$ is wrong.




      For $x ge 1$, $y ge 1$ you can square the inequality (since both
      sides are non-negative):
      $$
      sqrt{mathstrut x-1}+sqrt{mathstrut y-1}leq sqrt{mathstrut xy} \
      Longleftrightarrow (x-1) + (y-1) + 2 sqrt{mathstrut x-1}sqrt{mathstrut y-1} le xy \
      Longleftrightarrow 0 le xy - x - y + 2 - 2 sqrt{mathstrut (x-1)(y-1)} \
      Longleftrightarrow 0 le (x-1)(y-1) - 2 sqrt{mathstrut (x-1)(y-1)} + 1
      $$
      With $t := sqrt{(x-1)(y-1)}$ the right-hand side is
      $$
      t^2 - 2 t + 1 = (t-1)^2 ge 0 , .
      $$
      so that the inequality is true.



      It follows also that equality holds if and only if $t = 1$,
      i.e. if $(x-1)(y-1) = 1$.






      share|cite|improve this answer











      $endgroup$



      There are some errors in your calculation, e.g. a missing factor 2 in
      $$
      (sqrt{x-1}+sqrt{y-1})^2 = x - 1 + y - 1 + 2sqrt{x-1}sqrt{y-1}
      $$
      and in the last step the inequality sign is in the wrong direction and
      the number $3$ is wrong.




      For $x ge 1$, $y ge 1$ you can square the inequality (since both
      sides are non-negative):
      $$
      sqrt{mathstrut x-1}+sqrt{mathstrut y-1}leq sqrt{mathstrut xy} \
      Longleftrightarrow (x-1) + (y-1) + 2 sqrt{mathstrut x-1}sqrt{mathstrut y-1} le xy \
      Longleftrightarrow 0 le xy - x - y + 2 - 2 sqrt{mathstrut (x-1)(y-1)} \
      Longleftrightarrow 0 le (x-1)(y-1) - 2 sqrt{mathstrut (x-1)(y-1)} + 1
      $$
      With $t := sqrt{(x-1)(y-1)}$ the right-hand side is
      $$
      t^2 - 2 t + 1 = (t-1)^2 ge 0 , .
      $$
      so that the inequality is true.



      It follows also that equality holds if and only if $t = 1$,
      i.e. if $(x-1)(y-1) = 1$.







      share|cite|improve this answer














      share|cite|improve this answer



      share|cite|improve this answer








      edited Nov 8 '15 at 22:30

























      answered Nov 8 '15 at 21:23









      Martin RMartin R

      27.8k33255




      27.8k33255












      • $begingroup$
        Thank you so much for the answer ! :)
        $endgroup$
        – user233658
        Nov 8 '15 at 21:38










      • $begingroup$
        @user233658: You are welcome!
        $endgroup$
        – Martin R
        Nov 8 '15 at 21:43


















      • $begingroup$
        Thank you so much for the answer ! :)
        $endgroup$
        – user233658
        Nov 8 '15 at 21:38










      • $begingroup$
        @user233658: You are welcome!
        $endgroup$
        – Martin R
        Nov 8 '15 at 21:43
















      $begingroup$
      Thank you so much for the answer ! :)
      $endgroup$
      – user233658
      Nov 8 '15 at 21:38




      $begingroup$
      Thank you so much for the answer ! :)
      $endgroup$
      – user233658
      Nov 8 '15 at 21:38












      $begingroup$
      @user233658: You are welcome!
      $endgroup$
      – Martin R
      Nov 8 '15 at 21:43




      $begingroup$
      @user233658: You are welcome!
      $endgroup$
      – Martin R
      Nov 8 '15 at 21:43











      1












      $begingroup$

      Starting with “Assuming that $X$” when $X$ is the thing to be proved is not the right thing to do. You can make your computations easier if you set
      $$
      t=sqrt{x-1},quad u=sqrt{y-1}
      $$
      so $x=t^2+1$ and $y=u^2+1$. The inequality to be proved becomes
      $$
      t+ulesqrt{(t^2+1)(u^2+1)}
      $$
      that has a single radical. Since $t+uge0$, the inequality is equivalent to
      $$
      (t+u)^2le(t^2+1)(u^2+1)
      $$
      that is,
      $$
      t^2+2tu+u^2le t^2u^2+t^2+u^2+1
      $$
      and so equivalent to
      $$
      0le t^2u^2-2tu+1
      $$
      (by transporting terms to the right-hand side) which becomes
      $$
      0le(tu-1)^2
      $$
      which is true.



      Don't forget the $2$ when you square! You basically did $(a+b)^2=a^2+ab+b^2$ and this is where you got stuck.






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        Starting with “Assuming that $X$” when $X$ is the thing to be proved is not the right thing to do. You can make your computations easier if you set
        $$
        t=sqrt{x-1},quad u=sqrt{y-1}
        $$
        so $x=t^2+1$ and $y=u^2+1$. The inequality to be proved becomes
        $$
        t+ulesqrt{(t^2+1)(u^2+1)}
        $$
        that has a single radical. Since $t+uge0$, the inequality is equivalent to
        $$
        (t+u)^2le(t^2+1)(u^2+1)
        $$
        that is,
        $$
        t^2+2tu+u^2le t^2u^2+t^2+u^2+1
        $$
        and so equivalent to
        $$
        0le t^2u^2-2tu+1
        $$
        (by transporting terms to the right-hand side) which becomes
        $$
        0le(tu-1)^2
        $$
        which is true.



        Don't forget the $2$ when you square! You basically did $(a+b)^2=a^2+ab+b^2$ and this is where you got stuck.






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          Starting with “Assuming that $X$” when $X$ is the thing to be proved is not the right thing to do. You can make your computations easier if you set
          $$
          t=sqrt{x-1},quad u=sqrt{y-1}
          $$
          so $x=t^2+1$ and $y=u^2+1$. The inequality to be proved becomes
          $$
          t+ulesqrt{(t^2+1)(u^2+1)}
          $$
          that has a single radical. Since $t+uge0$, the inequality is equivalent to
          $$
          (t+u)^2le(t^2+1)(u^2+1)
          $$
          that is,
          $$
          t^2+2tu+u^2le t^2u^2+t^2+u^2+1
          $$
          and so equivalent to
          $$
          0le t^2u^2-2tu+1
          $$
          (by transporting terms to the right-hand side) which becomes
          $$
          0le(tu-1)^2
          $$
          which is true.



          Don't forget the $2$ when you square! You basically did $(a+b)^2=a^2+ab+b^2$ and this is where you got stuck.






          share|cite|improve this answer









          $endgroup$



          Starting with “Assuming that $X$” when $X$ is the thing to be proved is not the right thing to do. You can make your computations easier if you set
          $$
          t=sqrt{x-1},quad u=sqrt{y-1}
          $$
          so $x=t^2+1$ and $y=u^2+1$. The inequality to be proved becomes
          $$
          t+ulesqrt{(t^2+1)(u^2+1)}
          $$
          that has a single radical. Since $t+uge0$, the inequality is equivalent to
          $$
          (t+u)^2le(t^2+1)(u^2+1)
          $$
          that is,
          $$
          t^2+2tu+u^2le t^2u^2+t^2+u^2+1
          $$
          and so equivalent to
          $$
          0le t^2u^2-2tu+1
          $$
          (by transporting terms to the right-hand side) which becomes
          $$
          0le(tu-1)^2
          $$
          which is true.



          Don't forget the $2$ when you square! You basically did $(a+b)^2=a^2+ab+b^2$ and this is where you got stuck.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 8 '15 at 22:01









          egregegreg

          180k1485202




          180k1485202























              0












              $begingroup$

              WLOG $sqrt x=sec A,sqrt y=sec B$ where $0<A,B<dfracpi2$



              $sqrt{x-1}+sqrt{y-1}=tan A+tan B=dfrac{sin(A+B)}{cos Acos B}lesqrt{sec Asec B}=?$






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                WLOG $sqrt x=sec A,sqrt y=sec B$ where $0<A,B<dfracpi2$



                $sqrt{x-1}+sqrt{y-1}=tan A+tan B=dfrac{sin(A+B)}{cos Acos B}lesqrt{sec Asec B}=?$






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  WLOG $sqrt x=sec A,sqrt y=sec B$ where $0<A,B<dfracpi2$



                  $sqrt{x-1}+sqrt{y-1}=tan A+tan B=dfrac{sin(A+B)}{cos Acos B}lesqrt{sec Asec B}=?$






                  share|cite|improve this answer









                  $endgroup$



                  WLOG $sqrt x=sec A,sqrt y=sec B$ where $0<A,B<dfracpi2$



                  $sqrt{x-1}+sqrt{y-1}=tan A+tan B=dfrac{sin(A+B)}{cos Acos B}lesqrt{sec Asec B}=?$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 1 at 4:03









                  lab bhattacharjeelab bhattacharjee

                  224k15156274




                  224k15156274






























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                      Human spaceflight

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                      Human spaceflight From Wikipedia, the free encyclopedia Jump to navigation Jump to search "Space traveler" redirects here. For other uses, see Space traveler (disambiguation). @media all and (max-width:720px){.mw-parser-output .tmulti>.thumbinner{width:100%!important;max-width:none!important}.mw-parser-output .tmulti .tsingle{float:none!important;max-width:none!important;width:100%!important;text-align:center}} Apollo 11 crewmember Buzz Aldrin on the Moon, 1969 International Space Station crewmember Tracy Caldwell Dyson views the Earth, 2010 Space Shuttle Discovery heads into space with a crew aboard, STS-121 in 2006 Inside a space suit on the Canadarm, 1993 Human spaceflight (also referred to as crewed spaceflight or manned spaceflight ) is space travel with a crew or passengers aboard the spacecraft. Spacecraft carrying people may be operated directly, by human crew, or it may be e...
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                      Can not write log (Is /dev/pts mounted?) - openpty in Ubuntu-on-Windows?

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                      0 1 What exactly does the error message beginning with E: mean? I assume it has to do with the file structure on Ubuntu-on-Windows, but what exactly? The following package was automatically installed and is no longer required: os-prober Use 'apt-get autoremove' to remove it. The following extra packages will be installed: libxslt1.1 The following NEW packages will be installed: libxslt1.1 xmlstarlet 0 upgraded, 2 newly installed, 0 to remove and 0 not upgraded. Need to get 435 kB of archives. After this operation, 1023 kB of additional disk space will be used. Get:1 http://archive.ubuntu.com/ubuntu/ trusty/main libxslt1.1 amd64 1.1.28-2build1 [145 kB] Get:2 http://archive.ubuntu.com/ubuntu/ trusty/universe xmlstarlet amd64 1.5.0-1 [290 kB] ...
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                      張江高科駅

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                      張江高科駅 ちょうこうこうか Zhangjiang Hi-Tech Park 所在地 中国上海市浦東新区 所属事業者 上海地鉄運営有限公司（上海軌道交通・駅詳細） 上海浦東現代有軌交通有限公司（駅詳細） テンプレートを表示 張江高科駅 各種表記 繁体字： 張江高科站 簡体字： 张江高科站 拼音 ： Zhàngjiāng Gāokē Zhàn 発音： ヂャンジャン ガオクー ヂャン 英文： Zhangjiang Hi-Tech Park Station テンプレートを表示 張江高科駅 （ちょうこうこうか-えき）は中華人民共和国上海市浦東新区に位置する、上海軌道交通2号線の駅である。 なお、張江高科駅前にある張江軌道電車の 張江地鉄駅駅 についてもここで述べる。 目次 1 利用可能な鉄道路線 2 沿革 3 駅構造 3.1 上海軌道交通 3.2 張江軌道電車 4 駅周辺 5 隣の駅 6 注釈 7 関連項目 利用可能な鉄道路線 上海軌道交通 ■ 2号線 上海浦東現代有軌交通有限公司 ■ 張江軌道電車 沿革 高架駅時代の姿 2000年12月27日 - 2号線の駅が高架駅として開業。 2010年 1月1日 - 張江軌道電車の開業に伴い、張江地鉄駅駅（停留所）を設置。 2月13日 - 駅地下化に伴い高架駅の運用停止。 2月24日 - 10日間の閉鎖を経て地下新駅の開業。 2017年2月9日 - リニューアル工事を経て再開業。乗降客の増加に対応し、ホームへのエレベーター、エスカレーターを増設、券売機エリアを拡張した [1] 。 駅構造 上海軌道交通 上海軌道交通 張江高科駅 5号口 ちょうこうこうか Zhangjiang Hi-Tech Park ◄ 竜陽路 (2.60km) (1.61km) 金科路 ► 所在地 中国上海市浦東新区松涛路 祖冲之路 （中国語版） 北緯31度12分11秒 東経121度34分50秒  /  北緯31.2...
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