$x,yinBbb R,$ find the maxima of $fracx+2y+3sqrtx^2+y^2+1$Inequality.$sqrtfrac2ab+c+sqrtfrac2bc+a+sqrtfrac2ca+b leq 3$How find this inequality $sqrta^2+64+sqrtb^2+1$Inequality. $fracasqrtb+fracbsqrtc+fraccsqrtageq3$How find this minimum $a+b+c+sqrta^2+b^2+sqrtb^2+c^2+sqrta^2+c^2$Prove that $fracab+2c+3d+fracbc+2d+3a+fraccd+2a+3b+fracda+2b+3c geq frac23.$Find the Maximum value of $fracxsqrtx+y+fracysqrty+z+fraczsqrtz+x$Find the minimize value of $A=frac1sqrta^2+b^2+c^2+1-frac2(a+1)(b+1)(c+1)$Find this maximum of the $fracsqrt34x^2+fracsqrt(9-x^2)(x^2-1)4$Cauchy–Schwarz inequality to find minimum of a functionFind minimum value of $sum frac sqrt asqrt b +sqrt c-sqrt a$

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$x,yinBbb R,$ find the maxima of $fracx+2y+3sqrtx^2+y^2+1$


Inequality.$sqrtfrac2ab+c}+sqrtfrac2bc+a+sqrt{frac2ca+b leq 3$How find this inequality $sqrta^2+64+sqrtb^2+1$Inequality. $fracasqrtb+fracbsqrtc+fraccsqrtageq3$How find this minimum $a+b+c+sqrta^2+b^2+sqrtb^2+c^2+sqrta^2+c^2$Prove that $fracab+2c+3d+fracbc+2d+3a+fraccd+2a+3b+fracda+2b+3c geq frac23.$Find the Maximum value of $fracxsqrtx+y+fracysqrty+z+fraczsqrtz+x$Find the minimize value of $A=frac1sqrta^2+b^2+c^2+1-frac2(a+1)(b+1)(c+1)$Find this maximum of the $fracsqrt34x^2+fracsqrt(9-x^2)(x^2-1)4$Cauchy–Schwarz inequality to find minimum of a functionFind minimum value of $sum frac sqrt asqrt b +sqrt c-sqrt a$













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I want to use Cauchy–Schwarz inequality, I sqared $fracx+2y+3sqrtx^2+y^2+1$ and got $fracx^2+4y^2+9x^2+y^2+1$, not sure if I am doing fine






real-analysis inequality maxima-minima cauchy-schwarz-inequality







share|cite|improve this question











$endgroup$











  • $begingroup$
    Hint: One of the vectors should be $(x, y, 1)$.
    $endgroup$
    – PkT
    2 days ago






  • 2




    $begingroup$
    $x^2+4y^2+9$ is NOT the square of $x+2y+3$.
    $endgroup$
    – Crostul
    2 days ago










  • $begingroup$
    @Crostul Wow I didn't notice that serious mistake, thank you.
    $endgroup$
    – David
    2 days ago















0












$begingroup$


I want to use Cauchy–Schwarz inequality, I sqared $fracx+2y+3sqrtx^2+y^2+1$ and got $fracx^2+4y^2+9x^2+y^2+1$, not sure if I am doing fine






real-analysis inequality maxima-minima cauchy-schwarz-inequality







share|cite|improve this question











$endgroup$











  • $begingroup$
    Hint: One of the vectors should be $(x, y, 1)$.
    $endgroup$
    – PkT
    2 days ago






  • 2




    $begingroup$
    $x^2+4y^2+9$ is NOT the square of $x+2y+3$.
    $endgroup$
    – Crostul
    2 days ago










  • $begingroup$
    @Crostul Wow I didn't notice that serious mistake, thank you.
    $endgroup$
    – David
    2 days ago













0












0








0


2



$begingroup$


I want to use Cauchy–Schwarz inequality, I sqared $fracx+2y+3sqrtx^2+y^2+1$ and got $fracx^2+4y^2+9x^2+y^2+1$, not sure if I am doing fine






real-analysis inequality maxima-minima cauchy-schwarz-inequality







share|cite|improve this question











$endgroup$




I want to use Cauchy–Schwarz inequality, I sqared $fracx+2y+3sqrtx^2+y^2+1$ and got $fracx^2+4y^2+9x^2+y^2+1$, not sure if I am doing fine






real-analysis inequality maxima-minima cauchy-schwarz-inequality




real-analysis inequality maxima-minima cauchy-schwarz-inequality






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 2 days ago









Michael Rozenberg

107k1895199




107k1895199










asked 2 days ago









DavidDavid

454




454











  • $begingroup$
    Hint: One of the vectors should be $(x, y, 1)$.
    $endgroup$
    – PkT
    2 days ago






  • 2




    $begingroup$
    $x^2+4y^2+9$ is NOT the square of $x+2y+3$.
    $endgroup$
    – Crostul
    2 days ago










  • $begingroup$
    @Crostul Wow I didn't notice that serious mistake, thank you.
    $endgroup$
    – David
    2 days ago
















  • $begingroup$
    Hint: One of the vectors should be $(x, y, 1)$.
    $endgroup$
    – PkT
    2 days ago






  • 2




    $begingroup$
    $x^2+4y^2+9$ is NOT the square of $x+2y+3$.
    $endgroup$
    – Crostul
    2 days ago










  • $begingroup$
    @Crostul Wow I didn't notice that serious mistake, thank you.
    $endgroup$
    – David
    2 days ago















$begingroup$
Hint: One of the vectors should be $(x, y, 1)$.
$endgroup$
– PkT
2 days ago




$begingroup$
Hint: One of the vectors should be $(x, y, 1)$.
$endgroup$
– PkT
2 days ago




2




2




$begingroup$
$x^2+4y^2+9$ is NOT the square of $x+2y+3$.
$endgroup$
– Crostul
2 days ago




$begingroup$
$x^2+4y^2+9$ is NOT the square of $x+2y+3$.
$endgroup$
– Crostul
2 days ago












$begingroup$
@Crostul Wow I didn't notice that serious mistake, thank you.
$endgroup$
– David
2 days ago




$begingroup$
@Crostul Wow I didn't notice that serious mistake, thank you.
$endgroup$
– David
2 days ago










2 Answers
2






active

oldest

votes


















2












$begingroup$

Not that $$(x+2y+3)^2le (x^2+y^2+1)(1+4+9)$$






share|cite|improve this answer









$endgroup$




















    2












    $begingroup$

    By C-S
    $$fracx+2y+3sqrtx^2+y^2+1leqsqrtfrac14(x+2y+3)^2(1^2+2^2+3^2)(x^2+y^2+1)leq$$
    $$leqsqrtfrac14(x+2y+3)^2(x+2y+3)^2=sqrt14.$$
    The equality occurs for $(x,y,1)||(1,2,3)$ and $x+2y+3>0,$ id est, for $(x,y)=left(frac13,frac23right),$



    which says that we got a maximal value.






    share|cite|improve this answer









    $endgroup$












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      2 Answers
      2






      active

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      2 Answers
      2






      active

      oldest

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      active

      oldest

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      active

      oldest

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      2












      $begingroup$

      Not that $$(x+2y+3)^2le (x^2+y^2+1)(1+4+9)$$






      share|cite|improve this answer









      $endgroup$

















        2












        $begingroup$

        Not that $$(x+2y+3)^2le (x^2+y^2+1)(1+4+9)$$






        share|cite|improve this answer









        $endgroup$















          2












          2








          2





          $begingroup$

          Not that $$(x+2y+3)^2le (x^2+y^2+1)(1+4+9)$$






          share|cite|improve this answer









          $endgroup$



          Not that $$(x+2y+3)^2le (x^2+y^2+1)(1+4+9)$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 2 days ago









          Dr. Sonnhard GraubnerDr. Sonnhard Graubner

          77.7k42866




          77.7k42866





















              2












              $begingroup$

              By C-S
              $$fracx+2y+3sqrtx^2+y^2+1leqsqrtfrac14(x+2y+3)^2(1^2+2^2+3^2)(x^2+y^2+1)leq$$
              $$leqsqrtfrac14(x+2y+3)^2(x+2y+3)^2=sqrt14.$$
              The equality occurs for $(x,y,1)||(1,2,3)$ and $x+2y+3>0,$ id est, for $(x,y)=left(frac13,frac23right),$



              which says that we got a maximal value.






              share|cite|improve this answer









              $endgroup$

















                2












                $begingroup$

                By C-S
                $$fracx+2y+3sqrtx^2+y^2+1leqsqrtfrac14(x+2y+3)^2(1^2+2^2+3^2)(x^2+y^2+1)leq$$
                $$leqsqrtfrac14(x+2y+3)^2(x+2y+3)^2=sqrt14.$$
                The equality occurs for $(x,y,1)||(1,2,3)$ and $x+2y+3>0,$ id est, for $(x,y)=left(frac13,frac23right),$



                which says that we got a maximal value.






                share|cite|improve this answer









                $endgroup$















                  2












                  2








                  2





                  $begingroup$

                  By C-S
                  $$fracx+2y+3sqrtx^2+y^2+1leqsqrtfrac14(x+2y+3)^2(1^2+2^2+3^2)(x^2+y^2+1)leq$$
                  $$leqsqrtfrac14(x+2y+3)^2(x+2y+3)^2=sqrt14.$$
                  The equality occurs for $(x,y,1)||(1,2,3)$ and $x+2y+3>0,$ id est, for $(x,y)=left(frac13,frac23right),$



                  which says that we got a maximal value.






                  share|cite|improve this answer









                  $endgroup$



                  By C-S
                  $$fracx+2y+3sqrtx^2+y^2+1leqsqrtfrac14(x+2y+3)^2(1^2+2^2+3^2)(x^2+y^2+1)leq$$
                  $$leqsqrtfrac14(x+2y+3)^2(x+2y+3)^2=sqrt14.$$
                  The equality occurs for $(x,y,1)||(1,2,3)$ and $x+2y+3>0,$ id est, for $(x,y)=left(frac13,frac23right),$



                  which says that we got a maximal value.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 2 days ago









                  Michael RozenbergMichael Rozenberg

                  107k1895199




                  107k1895199



























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