Inequality with $ax^2+bx+c$












0












$begingroup$


Let $a,b,c,x,y > 0$ reals prove that:
$$(ax^2+bx+c)(ay^2-by+c) geq (4ac-b^2)xy$$
What I have done is this:
$$ax^2+bx+c=a left (x+frac{b}{2a} right)^2+frac{4ac-b^2}{4a}$$
$$ay^2-by+c=a left (y-frac{b}{2a} right)^2+frac{4ac-b^2}{4a}$$
From here I have two cases:
$ 4ac-b^2 >0$
$$left[ sqrt{a}^2 left (x+frac{b}{2a} right)^2+sqrt frac{4ac-b^2}{4a}^2right ]cdot left[ sqrt{a}^2 left (y-frac{b}{2a} right)^2+sqrt frac{4ac-b^2}{4a}^2right ]geq$$
$$left [ sqrt a left (x+frac{b}{2a} right)cdotsqrt frac{4ac-b^2}{4a}+ sqrt a left (y-frac{b}{2a} right)cdotsqrt frac{4ac-b^2}{4a}right ]^2$$
This is just Cauchy.
$$left[ sqrt{a}^2 left (x+frac{b}{2a} right)^2+sqrt frac{4ac-b^2}{4a}^2right ]cdot left[ sqrt{a}^2 left (y-frac{b}{2a} right)^2+sqrt frac{4ac-b^2}{4a}^2right ]geq$$
$$a cdot frac{4ac-b^2}{4a} left ( x+frac{b}{2a}+y-frac{b}{2a} right )^2=(4ac-b^2) left ( frac{x+y}{2}right )^2geq (4ac-b^2)xy. $$
Now the second case I have trouble proving : $4ac-b^2 < 0$.
If someone can take a look and give me a solution or a hint I would much appreciate.






inequality cauchy-schwarz-inequality







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    0












    $begingroup$


    Let $a,b,c,x,y > 0$ reals prove that:
    $$(ax^2+bx+c)(ay^2-by+c) geq (4ac-b^2)xy$$
    What I have done is this:
    $$ax^2+bx+c=a left (x+frac{b}{2a} right)^2+frac{4ac-b^2}{4a}$$
    $$ay^2-by+c=a left (y-frac{b}{2a} right)^2+frac{4ac-b^2}{4a}$$
    From here I have two cases:
    $ 4ac-b^2 >0$
    $$left[ sqrt{a}^2 left (x+frac{b}{2a} right)^2+sqrt frac{4ac-b^2}{4a}^2right ]cdot left[ sqrt{a}^2 left (y-frac{b}{2a} right)^2+sqrt frac{4ac-b^2}{4a}^2right ]geq$$
    $$left [ sqrt a left (x+frac{b}{2a} right)cdotsqrt frac{4ac-b^2}{4a}+ sqrt a left (y-frac{b}{2a} right)cdotsqrt frac{4ac-b^2}{4a}right ]^2$$
    This is just Cauchy.
    $$left[ sqrt{a}^2 left (x+frac{b}{2a} right)^2+sqrt frac{4ac-b^2}{4a}^2right ]cdot left[ sqrt{a}^2 left (y-frac{b}{2a} right)^2+sqrt frac{4ac-b^2}{4a}^2right ]geq$$
    $$a cdot frac{4ac-b^2}{4a} left ( x+frac{b}{2a}+y-frac{b}{2a} right )^2=(4ac-b^2) left ( frac{x+y}{2}right )^2geq (4ac-b^2)xy. $$
    Now the second case I have trouble proving : $4ac-b^2 < 0$.
    If someone can take a look and give me a solution or a hint I would much appreciate.






    inequality cauchy-schwarz-inequality







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      Let $a,b,c,x,y > 0$ reals prove that:
      $$(ax^2+bx+c)(ay^2-by+c) geq (4ac-b^2)xy$$
      What I have done is this:
      $$ax^2+bx+c=a left (x+frac{b}{2a} right)^2+frac{4ac-b^2}{4a}$$
      $$ay^2-by+c=a left (y-frac{b}{2a} right)^2+frac{4ac-b^2}{4a}$$
      From here I have two cases:
      $ 4ac-b^2 >0$
      $$left[ sqrt{a}^2 left (x+frac{b}{2a} right)^2+sqrt frac{4ac-b^2}{4a}^2right ]cdot left[ sqrt{a}^2 left (y-frac{b}{2a} right)^2+sqrt frac{4ac-b^2}{4a}^2right ]geq$$
      $$left [ sqrt a left (x+frac{b}{2a} right)cdotsqrt frac{4ac-b^2}{4a}+ sqrt a left (y-frac{b}{2a} right)cdotsqrt frac{4ac-b^2}{4a}right ]^2$$
      This is just Cauchy.
      $$left[ sqrt{a}^2 left (x+frac{b}{2a} right)^2+sqrt frac{4ac-b^2}{4a}^2right ]cdot left[ sqrt{a}^2 left (y-frac{b}{2a} right)^2+sqrt frac{4ac-b^2}{4a}^2right ]geq$$
      $$a cdot frac{4ac-b^2}{4a} left ( x+frac{b}{2a}+y-frac{b}{2a} right )^2=(4ac-b^2) left ( frac{x+y}{2}right )^2geq (4ac-b^2)xy. $$
      Now the second case I have trouble proving : $4ac-b^2 < 0$.
      If someone can take a look and give me a solution or a hint I would much appreciate.






      inequality cauchy-schwarz-inequality







      share|cite|improve this question









      $endgroup$




      Let $a,b,c,x,y > 0$ reals prove that:
      $$(ax^2+bx+c)(ay^2-by+c) geq (4ac-b^2)xy$$
      What I have done is this:
      $$ax^2+bx+c=a left (x+frac{b}{2a} right)^2+frac{4ac-b^2}{4a}$$
      $$ay^2-by+c=a left (y-frac{b}{2a} right)^2+frac{4ac-b^2}{4a}$$
      From here I have two cases:
      $ 4ac-b^2 >0$
      $$left[ sqrt{a}^2 left (x+frac{b}{2a} right)^2+sqrt frac{4ac-b^2}{4a}^2right ]cdot left[ sqrt{a}^2 left (y-frac{b}{2a} right)^2+sqrt frac{4ac-b^2}{4a}^2right ]geq$$
      $$left [ sqrt a left (x+frac{b}{2a} right)cdotsqrt frac{4ac-b^2}{4a}+ sqrt a left (y-frac{b}{2a} right)cdotsqrt frac{4ac-b^2}{4a}right ]^2$$
      This is just Cauchy.
      $$left[ sqrt{a}^2 left (x+frac{b}{2a} right)^2+sqrt frac{4ac-b^2}{4a}^2right ]cdot left[ sqrt{a}^2 left (y-frac{b}{2a} right)^2+sqrt frac{4ac-b^2}{4a}^2right ]geq$$
      $$a cdot frac{4ac-b^2}{4a} left ( x+frac{b}{2a}+y-frac{b}{2a} right )^2=(4ac-b^2) left ( frac{x+y}{2}right )^2geq (4ac-b^2)xy. $$
      Now the second case I have trouble proving : $4ac-b^2 < 0$.
      If someone can take a look and give me a solution or a hint I would much appreciate.






      inequality cauchy-schwarz-inequality




      inequality cauchy-schwarz-inequality






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      asked Jan 23 at 17:24









      mathlearningmathlearning

      1787




      1787






















          2 Answers
          2






          active

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          1












          $begingroup$

          Hint: Use AM-GM inequality to obtain
          $$
          ax+b+frac{c}{x}ge b+2sqrt{ac},
          $$
          $$
          ay-b+frac{c}{y}ge -b+2sqrt{ac}.
          $$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Yes. I have tried this too. But it's true only if $4ac-b^2>0$. And I don't know what to do if $4ac-b^2<0$
            $endgroup$
            – mathlearning
            Jan 23 at 17:31






          • 1




            $begingroup$
            @ovetz13 Umm... the given inequality seems to fail if $4ac-b^2< 0$. Look at the example: $(a,b,c)=(1,3,1)$. Then for $(x,y)=(3,1)$, the LHS becomes $-19$ while the RHS is $-15$. It fails to hold $-19geqslant -15$.
            $endgroup$
            – Song
            Jan 23 at 17:45












          • $begingroup$
            So I can say that the given inequality is true only if $4ac-b^2>0$. Problem solved I guess.
            $endgroup$
            – mathlearning
            Jan 23 at 17:47










          • $begingroup$
            @ovetz13 Yep, I think so.
            $endgroup$
            – Song
            Jan 23 at 17:48



















          1












          $begingroup$

          It's wrong.



          Try $b=c=1$, $x=1$, $y=3$ and $a=frac{1}{3}.$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I know. Thank you.
            $endgroup$
            – mathlearning
            Jan 23 at 21:04











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          2 Answers
          2






          active

          oldest

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          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          1












          $begingroup$

          Hint: Use AM-GM inequality to obtain
          $$
          ax+b+frac{c}{x}ge b+2sqrt{ac},
          $$
          $$
          ay-b+frac{c}{y}ge -b+2sqrt{ac}.
          $$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Yes. I have tried this too. But it's true only if $4ac-b^2>0$. And I don't know what to do if $4ac-b^2<0$
            $endgroup$
            – mathlearning
            Jan 23 at 17:31






          • 1




            $begingroup$
            @ovetz13 Umm... the given inequality seems to fail if $4ac-b^2< 0$. Look at the example: $(a,b,c)=(1,3,1)$. Then for $(x,y)=(3,1)$, the LHS becomes $-19$ while the RHS is $-15$. It fails to hold $-19geqslant -15$.
            $endgroup$
            – Song
            Jan 23 at 17:45












          • $begingroup$
            So I can say that the given inequality is true only if $4ac-b^2>0$. Problem solved I guess.
            $endgroup$
            – mathlearning
            Jan 23 at 17:47










          • $begingroup$
            @ovetz13 Yep, I think so.
            $endgroup$
            – Song
            Jan 23 at 17:48
















          1












          $begingroup$

          Hint: Use AM-GM inequality to obtain
          $$
          ax+b+frac{c}{x}ge b+2sqrt{ac},
          $$
          $$
          ay-b+frac{c}{y}ge -b+2sqrt{ac}.
          $$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Yes. I have tried this too. But it's true only if $4ac-b^2>0$. And I don't know what to do if $4ac-b^2<0$
            $endgroup$
            – mathlearning
            Jan 23 at 17:31






          • 1




            $begingroup$
            @ovetz13 Umm... the given inequality seems to fail if $4ac-b^2< 0$. Look at the example: $(a,b,c)=(1,3,1)$. Then for $(x,y)=(3,1)$, the LHS becomes $-19$ while the RHS is $-15$. It fails to hold $-19geqslant -15$.
            $endgroup$
            – Song
            Jan 23 at 17:45












          • $begingroup$
            So I can say that the given inequality is true only if $4ac-b^2>0$. Problem solved I guess.
            $endgroup$
            – mathlearning
            Jan 23 at 17:47










          • $begingroup$
            @ovetz13 Yep, I think so.
            $endgroup$
            – Song
            Jan 23 at 17:48














          1












          1








          1





          $begingroup$

          Hint: Use AM-GM inequality to obtain
          $$
          ax+b+frac{c}{x}ge b+2sqrt{ac},
          $$
          $$
          ay-b+frac{c}{y}ge -b+2sqrt{ac}.
          $$






          share|cite|improve this answer









          $endgroup$



          Hint: Use AM-GM inequality to obtain
          $$
          ax+b+frac{c}{x}ge b+2sqrt{ac},
          $$
          $$
          ay-b+frac{c}{y}ge -b+2sqrt{ac}.
          $$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 23 at 17:29









          SongSong

          17.4k21246




          17.4k21246












          • $begingroup$
            Yes. I have tried this too. But it's true only if $4ac-b^2>0$. And I don't know what to do if $4ac-b^2<0$
            $endgroup$
            – mathlearning
            Jan 23 at 17:31






          • 1




            $begingroup$
            @ovetz13 Umm... the given inequality seems to fail if $4ac-b^2< 0$. Look at the example: $(a,b,c)=(1,3,1)$. Then for $(x,y)=(3,1)$, the LHS becomes $-19$ while the RHS is $-15$. It fails to hold $-19geqslant -15$.
            $endgroup$
            – Song
            Jan 23 at 17:45












          • $begingroup$
            So I can say that the given inequality is true only if $4ac-b^2>0$. Problem solved I guess.
            $endgroup$
            – mathlearning
            Jan 23 at 17:47










          • $begingroup$
            @ovetz13 Yep, I think so.
            $endgroup$
            – Song
            Jan 23 at 17:48


















          • $begingroup$
            Yes. I have tried this too. But it's true only if $4ac-b^2>0$. And I don't know what to do if $4ac-b^2<0$
            $endgroup$
            – mathlearning
            Jan 23 at 17:31






          • 1




            $begingroup$
            @ovetz13 Umm... the given inequality seems to fail if $4ac-b^2< 0$. Look at the example: $(a,b,c)=(1,3,1)$. Then for $(x,y)=(3,1)$, the LHS becomes $-19$ while the RHS is $-15$. It fails to hold $-19geqslant -15$.
            $endgroup$
            – Song
            Jan 23 at 17:45












          • $begingroup$
            So I can say that the given inequality is true only if $4ac-b^2>0$. Problem solved I guess.
            $endgroup$
            – mathlearning
            Jan 23 at 17:47










          • $begingroup$
            @ovetz13 Yep, I think so.
            $endgroup$
            – Song
            Jan 23 at 17:48
















          $begingroup$
          Yes. I have tried this too. But it's true only if $4ac-b^2>0$. And I don't know what to do if $4ac-b^2<0$
          $endgroup$
          – mathlearning
          Jan 23 at 17:31




          $begingroup$
          Yes. I have tried this too. But it's true only if $4ac-b^2>0$. And I don't know what to do if $4ac-b^2<0$
          $endgroup$
          – mathlearning
          Jan 23 at 17:31




          1




          1




          $begingroup$
          @ovetz13 Umm... the given inequality seems to fail if $4ac-b^2< 0$. Look at the example: $(a,b,c)=(1,3,1)$. Then for $(x,y)=(3,1)$, the LHS becomes $-19$ while the RHS is $-15$. It fails to hold $-19geqslant -15$.
          $endgroup$
          – Song
          Jan 23 at 17:45






          $begingroup$
          @ovetz13 Umm... the given inequality seems to fail if $4ac-b^2< 0$. Look at the example: $(a,b,c)=(1,3,1)$. Then for $(x,y)=(3,1)$, the LHS becomes $-19$ while the RHS is $-15$. It fails to hold $-19geqslant -15$.
          $endgroup$
          – Song
          Jan 23 at 17:45














          $begingroup$
          So I can say that the given inequality is true only if $4ac-b^2>0$. Problem solved I guess.
          $endgroup$
          – mathlearning
          Jan 23 at 17:47




          $begingroup$
          So I can say that the given inequality is true only if $4ac-b^2>0$. Problem solved I guess.
          $endgroup$
          – mathlearning
          Jan 23 at 17:47












          $begingroup$
          @ovetz13 Yep, I think so.
          $endgroup$
          – Song
          Jan 23 at 17:48




          $begingroup$
          @ovetz13 Yep, I think so.
          $endgroup$
          – Song
          Jan 23 at 17:48











          1












          $begingroup$

          It's wrong.



          Try $b=c=1$, $x=1$, $y=3$ and $a=frac{1}{3}.$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I know. Thank you.
            $endgroup$
            – mathlearning
            Jan 23 at 21:04
















          1












          $begingroup$

          It's wrong.



          Try $b=c=1$, $x=1$, $y=3$ and $a=frac{1}{3}.$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I know. Thank you.
            $endgroup$
            – mathlearning
            Jan 23 at 21:04














          1












          1








          1





          $begingroup$

          It's wrong.



          Try $b=c=1$, $x=1$, $y=3$ and $a=frac{1}{3}.$






          share|cite|improve this answer









          $endgroup$



          It's wrong.



          Try $b=c=1$, $x=1$, $y=3$ and $a=frac{1}{3}.$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 23 at 21:03









          Michael RozenbergMichael Rozenberg

          108k1895200




          108k1895200












          • $begingroup$
            I know. Thank you.
            $endgroup$
            – mathlearning
            Jan 23 at 21:04


















          • $begingroup$
            I know. Thank you.
            $endgroup$
            – mathlearning
            Jan 23 at 21:04
















          $begingroup$
          I know. Thank you.
          $endgroup$
          – mathlearning
          Jan 23 at 21:04




          $begingroup$
          I know. Thank you.
          $endgroup$
          – mathlearning
          Jan 23 at 21:04


















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