Proving a Cauchy-Schwarz-like inequality












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For real numbers $x_1,ldots,x_n$ and $y_1,ldots,y_n$ with $$x_1,y_1>0,
x_1^2>x_2^2+ldots+x_n^2, y_1^2>y_2^2+ldots+y_n^2,$$ show that
$$x_1y_1-x_2y_2-ldots-x_ny_n geq sqrt{(x_1^2-x_2^2-ldots-x_n^2)(y_1^2-y_2^2-ldots-y_n^2)}$$
and determine necessary and sufficient conditions for equality to hold.




So this inequality (perhaps obviously) comes from Lorentzian linear algebra. In particular, the Cauchy-Schwarz inequality for the Lorentz norm. The proof I saw actually simplified the problem considerably by assuming without loss of generality that $y=(1,0,0,ldots,0)$ (if I recall correctly). This 'wlog' was justified by a lemma we had proven earlier, that the positive Lorentz group acts transitively on $m$-dimensional time-like subspaces of $n$-dimensional hyperbolic space, which itself was somewhat involved, using Gram-Schmidt orthonormalisation in the proof.



That said, I would like to know whether there are any direct algebraic proofs of this, since in itself it is just an elementary algebra problem.I would be surprised if the properties of hyperbolic isometries were really necessary just to prove this inequality. I hope I haven't made a mistake in the problem statement.






inequality hyperbolic-geometry cauchy-schwarz-inequality







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    2












    $begingroup$



    For real numbers $x_1,ldots,x_n$ and $y_1,ldots,y_n$ with $$x_1,y_1>0,
    x_1^2>x_2^2+ldots+x_n^2, y_1^2>y_2^2+ldots+y_n^2,$$     show that
    $$x_1y_1-x_2y_2-ldots-x_ny_n geq sqrt{(x_1^2-x_2^2-ldots-x_n^2)(y_1^2-y_2^2-ldots-y_n^2)}$$
    and determine necessary and sufficient conditions for equality to hold.




    So this inequality (perhaps obviously) comes from Lorentzian linear algebra. In particular, the Cauchy-Schwarz inequality for the Lorentz norm. The proof I saw actually simplified the problem considerably by assuming without loss of generality that $y=(1,0,0,ldots,0)$ (if I recall correctly). This 'wlog' was justified by a lemma we had proven earlier, that the positive Lorentz group acts transitively on $m$-dimensional time-like subspaces of $n$-dimensional hyperbolic space, which itself was somewhat involved, using Gram-Schmidt orthonormalisation in the proof.



    That said, I would like to know whether there are any direct algebraic proofs of this, since in itself it is just an elementary algebra problem.I would be surprised if the properties of hyperbolic isometries were really necessary just to prove this inequality. I hope I haven't made a mistake in the problem statement.






    inequality hyperbolic-geometry cauchy-schwarz-inequality







    share|cite|improve this question









    $endgroup$















      2












      2








      2





      $begingroup$



      For real numbers $x_1,ldots,x_n$ and $y_1,ldots,y_n$ with $$x_1,y_1>0,
      x_1^2>x_2^2+ldots+x_n^2, y_1^2>y_2^2+ldots+y_n^2,$$       show that
      $$x_1y_1-x_2y_2-ldots-x_ny_n geq sqrt{(x_1^2-x_2^2-ldots-x_n^2)(y_1^2-y_2^2-ldots-y_n^2)}$$
      and determine necessary and sufficient conditions for equality to hold.




      So this inequality (perhaps obviously) comes from Lorentzian linear algebra. In particular, the Cauchy-Schwarz inequality for the Lorentz norm. The proof I saw actually simplified the problem considerably by assuming without loss of generality that $y=(1,0,0,ldots,0)$ (if I recall correctly). This 'wlog' was justified by a lemma we had proven earlier, that the positive Lorentz group acts transitively on $m$-dimensional time-like subspaces of $n$-dimensional hyperbolic space, which itself was somewhat involved, using Gram-Schmidt orthonormalisation in the proof.



      That said, I would like to know whether there are any direct algebraic proofs of this, since in itself it is just an elementary algebra problem.I would be surprised if the properties of hyperbolic isometries were really necessary just to prove this inequality. I hope I haven't made a mistake in the problem statement.






      inequality hyperbolic-geometry cauchy-schwarz-inequality







      share|cite|improve this question









      $endgroup$





      For real numbers $x_1,ldots,x_n$ and $y_1,ldots,y_n$ with $$x_1,y_1>0,
      x_1^2>x_2^2+ldots+x_n^2, y_1^2>y_2^2+ldots+y_n^2,$$       show that
      $$x_1y_1-x_2y_2-ldots-x_ny_n geq sqrt{(x_1^2-x_2^2-ldots-x_n^2)(y_1^2-y_2^2-ldots-y_n^2)}$$
      and determine necessary and sufficient conditions for equality to hold.




      So this inequality (perhaps obviously) comes from Lorentzian linear algebra. In particular, the Cauchy-Schwarz inequality for the Lorentz norm. The proof I saw actually simplified the problem considerably by assuming without loss of generality that $y=(1,0,0,ldots,0)$ (if I recall correctly). This 'wlog' was justified by a lemma we had proven earlier, that the positive Lorentz group acts transitively on $m$-dimensional time-like subspaces of $n$-dimensional hyperbolic space, which itself was somewhat involved, using Gram-Schmidt orthonormalisation in the proof.



      That said, I would like to know whether there are any direct algebraic proofs of this, since in itself it is just an elementary algebra problem.I would be surprised if the properties of hyperbolic isometries were really necessary just to prove this inequality. I hope I haven't made a mistake in the problem statement.






      inequality hyperbolic-geometry cauchy-schwarz-inequality




      inequality hyperbolic-geometry cauchy-schwarz-inequality






      share|cite|improve this question













      share|cite|improve this question











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      asked Jan 2 at 23:12









      AlephNullAlephNull

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          Indeed there is a simple direct proof. Let $alpha =sqrt {x_2^{2}+x_3^{2}+cdots +x_n^{2}}$ and $beta =sqrt {y_2^{2}+y_3^{2}+cdots +y_n^{2}}$. The inequality reduces to $x_1y_1 -alpha beta geq sqrt {x_1^{2}-alpha^{2}} sqrt{y_1^{2}-beta^{2}}$ which reduces to $2x_1y_1alpha beta leq x_1^{2}beta ^{2}+y_1^{2}alpha^{2}$ which is obvious.






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          • $begingroup$
            That really is a lot simpler!
            $endgroup$
            – AlephNull
            Jan 2 at 23:37











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          $begingroup$

          Indeed there is a simple direct proof. Let $alpha =sqrt {x_2^{2}+x_3^{2}+cdots +x_n^{2}}$ and $beta =sqrt {y_2^{2}+y_3^{2}+cdots +y_n^{2}}$. The inequality reduces to $x_1y_1 -alpha beta geq sqrt {x_1^{2}-alpha^{2}} sqrt{y_1^{2}-beta^{2}}$ which reduces to $2x_1y_1alpha beta leq x_1^{2}beta ^{2}+y_1^{2}alpha^{2}$ which is obvious.






          share|cite|improve this answer









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          • $begingroup$
            That really is a lot simpler!
            $endgroup$
            – AlephNull
            Jan 2 at 23:37
















          3












          $begingroup$

          Indeed there is a simple direct proof. Let $alpha =sqrt {x_2^{2}+x_3^{2}+cdots +x_n^{2}}$ and $beta =sqrt {y_2^{2}+y_3^{2}+cdots +y_n^{2}}$. The inequality reduces to $x_1y_1 -alpha beta geq sqrt {x_1^{2}-alpha^{2}} sqrt{y_1^{2}-beta^{2}}$ which reduces to $2x_1y_1alpha beta leq x_1^{2}beta ^{2}+y_1^{2}alpha^{2}$ which is obvious.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            That really is a lot simpler!
            $endgroup$
            – AlephNull
            Jan 2 at 23:37














          3












          3








          3





          $begingroup$

          Indeed there is a simple direct proof. Let $alpha =sqrt {x_2^{2}+x_3^{2}+cdots +x_n^{2}}$ and $beta =sqrt {y_2^{2}+y_3^{2}+cdots +y_n^{2}}$. The inequality reduces to $x_1y_1 -alpha beta geq sqrt {x_1^{2}-alpha^{2}} sqrt{y_1^{2}-beta^{2}}$ which reduces to $2x_1y_1alpha beta leq x_1^{2}beta ^{2}+y_1^{2}alpha^{2}$ which is obvious.






          share|cite|improve this answer









          $endgroup$



          Indeed there is a simple direct proof. Let $alpha =sqrt {x_2^{2}+x_3^{2}+cdots +x_n^{2}}$ and $beta =sqrt {y_2^{2}+y_3^{2}+cdots +y_n^{2}}$. The inequality reduces to $x_1y_1 -alpha beta geq sqrt {x_1^{2}-alpha^{2}} sqrt{y_1^{2}-beta^{2}}$ which reduces to $2x_1y_1alpha beta leq x_1^{2}beta ^{2}+y_1^{2}alpha^{2}$ which is obvious.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 2 at 23:26









          Kavi Rama MurthyKavi Rama Murthy

          65.6k42766




          65.6k42766












          • $begingroup$
            That really is a lot simpler!
            $endgroup$
            – AlephNull
            Jan 2 at 23:37


















          • $begingroup$
            That really is a lot simpler!
            $endgroup$
            – AlephNull
            Jan 2 at 23:37
















          $begingroup$
          That really is a lot simpler!
          $endgroup$
          – AlephNull
          Jan 2 at 23:37




          $begingroup$
          That really is a lot simpler!
          $endgroup$
          – AlephNull
          Jan 2 at 23:37


















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