Elementary inequality proof $2|tx|leq |x^2|+|t^2|$












1












$begingroup$


I am trying to prove the following inequality for $xin mathbb{R}$ und $tin mathbb{C}$:
$$2|tx|leq |t^2|+|x^2|$$



I have for $t=a+bi$:
$$|t^2|=|a^2-b^2+aib|=sqrt{left(a^2+b^2right)^2}$$
and
$$2|tx|=2sqrt{x^2}sqrt{a^2+b^2}$$
reaching
$$2sqrt{x^2}sqrt{a^2+b^2}leqsqrt{left(a^2+b^2right)^2}+x^2 $$



How do I continue?






algebra-precalculus inequality complex-numbers







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    I am trying to prove the following inequality for $xin mathbb{R}$ und $tin mathbb{C}$:
    $$2|tx|leq |t^2|+|x^2|$$



    I have for $t=a+bi$:
    $$|t^2|=|a^2-b^2+aib|=sqrt{left(a^2+b^2right)^2}$$
    and
    $$2|tx|=2sqrt{x^2}sqrt{a^2+b^2}$$
    reaching
    $$2sqrt{x^2}sqrt{a^2+b^2}leqsqrt{left(a^2+b^2right)^2}+x^2 $$



    How do I continue?






    algebra-precalculus inequality complex-numbers







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      I am trying to prove the following inequality for $xin mathbb{R}$ und $tin mathbb{C}$:
      $$2|tx|leq |t^2|+|x^2|$$



      I have for $t=a+bi$:
      $$|t^2|=|a^2-b^2+aib|=sqrt{left(a^2+b^2right)^2}$$
      and
      $$2|tx|=2sqrt{x^2}sqrt{a^2+b^2}$$
      reaching
      $$2sqrt{x^2}sqrt{a^2+b^2}leqsqrt{left(a^2+b^2right)^2}+x^2 $$



      How do I continue?






      algebra-precalculus inequality complex-numbers







      share|cite|improve this question











      $endgroup$




      I am trying to prove the following inequality for $xin mathbb{R}$ und $tin mathbb{C}$:
      $$2|tx|leq |t^2|+|x^2|$$



      I have for $t=a+bi$:
      $$|t^2|=|a^2-b^2+aib|=sqrt{left(a^2+b^2right)^2}$$
      and
      $$2|tx|=2sqrt{x^2}sqrt{a^2+b^2}$$
      reaching
      $$2sqrt{x^2}sqrt{a^2+b^2}leqsqrt{left(a^2+b^2right)^2}+x^2 $$



      How do I continue?






      algebra-precalculus inequality complex-numbers




      algebra-precalculus inequality complex-numbers






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 30 at 15:57









      José Carlos Santos

      172k22132239




      172k22132239










      asked Jan 30 at 15:21









      orangeorange

      687316




      687316






















          2 Answers
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          $begingroup$

          $$lvert xrvert^2+lvert trvert^2-2lvert xtrvert=lvert xrvert^2+lvert trvert^2-2lvert xrvertlvert trvert=bigl(lvert xrvert-lvert trvertbigr)^2geqslant0.$$






          share|cite|improve this answer









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          • $begingroup$
            I want $|t^2|$ not norm squared. The norm isn't muktiolivative for complex numbers right?
            $endgroup$
            – orange
            Jan 30 at 18:54










          • $begingroup$
            Yes, it is: $(forall z,winmathbb{C}):lvert zwrvert=lvert zrvertlvert wrvert$.
            $endgroup$
            – José Carlos Santos
            Jan 30 at 18:55



















          1












          $begingroup$

          Hint: Write $$0le |x|^2+|t|^2-2|tx|$$ it is $$0le (|x|-|t|)^2$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Why is that correct
            $endgroup$
            – orange
            Jan 30 at 15:27












          Your Answer





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          2 Answers
          2






          active

          oldest

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          2 Answers
          2






          active

          oldest

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          active

          oldest

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          active

          oldest

          votes









          3












          $begingroup$

          $$lvert xrvert^2+lvert trvert^2-2lvert xtrvert=lvert xrvert^2+lvert trvert^2-2lvert xrvertlvert trvert=bigl(lvert xrvert-lvert trvertbigr)^2geqslant0.$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I want $|t^2|$ not norm squared. The norm isn't muktiolivative for complex numbers right?
            $endgroup$
            – orange
            Jan 30 at 18:54










          • $begingroup$
            Yes, it is: $(forall z,winmathbb{C}):lvert zwrvert=lvert zrvertlvert wrvert$.
            $endgroup$
            – José Carlos Santos
            Jan 30 at 18:55
















          3












          $begingroup$

          $$lvert xrvert^2+lvert trvert^2-2lvert xtrvert=lvert xrvert^2+lvert trvert^2-2lvert xrvertlvert trvert=bigl(lvert xrvert-lvert trvertbigr)^2geqslant0.$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I want $|t^2|$ not norm squared. The norm isn't muktiolivative for complex numbers right?
            $endgroup$
            – orange
            Jan 30 at 18:54










          • $begingroup$
            Yes, it is: $(forall z,winmathbb{C}):lvert zwrvert=lvert zrvertlvert wrvert$.
            $endgroup$
            – José Carlos Santos
            Jan 30 at 18:55














          3












          3








          3





          $begingroup$

          $$lvert xrvert^2+lvert trvert^2-2lvert xtrvert=lvert xrvert^2+lvert trvert^2-2lvert xrvertlvert trvert=bigl(lvert xrvert-lvert trvertbigr)^2geqslant0.$$






          share|cite|improve this answer









          $endgroup$



          $$lvert xrvert^2+lvert trvert^2-2lvert xtrvert=lvert xrvert^2+lvert trvert^2-2lvert xrvertlvert trvert=bigl(lvert xrvert-lvert trvertbigr)^2geqslant0.$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 30 at 15:27









          José Carlos SantosJosé Carlos Santos

          172k22132239




          172k22132239












          • $begingroup$
            I want $|t^2|$ not norm squared. The norm isn't muktiolivative for complex numbers right?
            $endgroup$
            – orange
            Jan 30 at 18:54










          • $begingroup$
            Yes, it is: $(forall z,winmathbb{C}):lvert zwrvert=lvert zrvertlvert wrvert$.
            $endgroup$
            – José Carlos Santos
            Jan 30 at 18:55


















          • $begingroup$
            I want $|t^2|$ not norm squared. The norm isn't muktiolivative for complex numbers right?
            $endgroup$
            – orange
            Jan 30 at 18:54










          • $begingroup$
            Yes, it is: $(forall z,winmathbb{C}):lvert zwrvert=lvert zrvertlvert wrvert$.
            $endgroup$
            – José Carlos Santos
            Jan 30 at 18:55
















          $begingroup$
          I want $|t^2|$ not norm squared. The norm isn't muktiolivative for complex numbers right?
          $endgroup$
          – orange
          Jan 30 at 18:54




          $begingroup$
          I want $|t^2|$ not norm squared. The norm isn't muktiolivative for complex numbers right?
          $endgroup$
          – orange
          Jan 30 at 18:54












          $begingroup$
          Yes, it is: $(forall z,winmathbb{C}):lvert zwrvert=lvert zrvertlvert wrvert$.
          $endgroup$
          – José Carlos Santos
          Jan 30 at 18:55




          $begingroup$
          Yes, it is: $(forall z,winmathbb{C}):lvert zwrvert=lvert zrvertlvert wrvert$.
          $endgroup$
          – José Carlos Santos
          Jan 30 at 18:55











          1












          $begingroup$

          Hint: Write $$0le |x|^2+|t|^2-2|tx|$$ it is $$0le (|x|-|t|)^2$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Why is that correct
            $endgroup$
            – orange
            Jan 30 at 15:27
















          1












          $begingroup$

          Hint: Write $$0le |x|^2+|t|^2-2|tx|$$ it is $$0le (|x|-|t|)^2$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Why is that correct
            $endgroup$
            – orange
            Jan 30 at 15:27














          1












          1








          1





          $begingroup$

          Hint: Write $$0le |x|^2+|t|^2-2|tx|$$ it is $$0le (|x|-|t|)^2$$






          share|cite|improve this answer











          $endgroup$



          Hint: Write $$0le |x|^2+|t|^2-2|tx|$$ it is $$0le (|x|-|t|)^2$$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 30 at 15:28

























          answered Jan 30 at 15:26









          Dr. Sonnhard GraubnerDr. Sonnhard Graubner

          78.4k42867




          78.4k42867












          • $begingroup$
            Why is that correct
            $endgroup$
            – orange
            Jan 30 at 15:27


















          • $begingroup$
            Why is that correct
            $endgroup$
            – orange
            Jan 30 at 15:27
















          $begingroup$
          Why is that correct
          $endgroup$
          – orange
          Jan 30 at 15:27




          $begingroup$
          Why is that correct
          $endgroup$
          – orange
          Jan 30 at 15:27


















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