$a,b,cinBbb R^+, x,y,zin Bbb R, $ show that $fracx^2a+fracy^2b+fracz^2c ge frac(x+y+z)^2a+b+c$Prove $sumlimits_i=1^nfraca_i^2b_i geq frac(sumlimits_i=1^na_i)^2sumlimits_i=1^nb_i$Harmonic mean: show $maxax,by ge frac1a+b(x+y)$, $a,b>1$, $x,yge 0$Proof of analogue of the Cauchy-Schwarz inequality for random variablesOlympiad inequality $fraca2a + b + fracb2b + c + fracc2c + a leq 1$.Prove Inequality with Cauchy SchwarzShow that $frac1x_1(x_1+1)+frac1x_2(x_2+1)+cdots+frac1x_n(x_n+1)gefracn2$ with $x_1x_2cdots x_n=1$Estimating the product of timelike vectors in $Bbb R^n_nu$Showing that two vectors are not orthogonal, using Cauchy-Schwarz$AM$- $QM$ inequalityCauchy-Schwarz inequality to prove $A$ is spd$x,yinBbb R,$ find the maxima of $fracx+2y+3sqrtx^2+y^2+1$

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$a,b,cinBbb R^+, x,y,zin Bbb R, $ show that $fracx^2a+fracy^2b+fracz^2c ge frac(x+y+z)^2a+b+c$


Prove $sumlimits_i=1^nfraca_i^2b_i geq frac(sumlimits_i=1^na_i)^2sumlimits_i=1^nb_i$Harmonic mean: show $maxax,by ge frac1a+b(x+y)$, $a,b>1$, $x,yge 0$Proof of analogue of the Cauchy-Schwarz inequality for random variablesOlympiad inequality $fraca2a + b + fracb2b + c + fracc2c + a leq 1$.Prove Inequality with Cauchy SchwarzShow that $frac1x_1(x_1+1)+frac1x_2(x_2+1)+cdots+frac1x_n(x_n+1)gefracn2$ with $x_1x_2cdots x_n=1$Estimating the product of timelike vectors in $Bbb R^n_nu$Showing that two vectors are not orthogonal, using Cauchy-Schwarz$AM$- $QM$ inequalityCauchy-Schwarz inequality to prove $A$ is spd$x,yinBbb R,$ find the maxima of $fracx+2y+3sqrtx^2+y^2+1$













1












$begingroup$


$a,b,cinBbb R^+, x,y,zin Bbb R, $ show that $fracx^2a+fracy^2b+fracz^2c ge frac(x+y+z)^2a+b+c$ (use Cauchy–Schwarz inequality)



I have trouble finding the two vectors. Is it $(x,y,z)$ and $(a,b,c)$?



I need a hint






inequality cauchy-schwarz-inequality







share|cite|improve this question











$endgroup$







  • 2




    $begingroup$
    This is known as Titus Lemma. More details here
    $endgroup$
    – rtybase
    2 days ago










  • $begingroup$
    Try $(x/sqrt a, y/sqrt b, z/sqrt c)$ and $(sqrt a, sqrt b, sqrt c)$…
    $endgroup$
    – Macavity
    2 days ago















1












$begingroup$


$a,b,cinBbb R^+, x,y,zin Bbb R, $ show that $fracx^2a+fracy^2b+fracz^2c ge frac(x+y+z)^2a+b+c$ (use Cauchy–Schwarz inequality)



I have trouble finding the two vectors. Is it $(x,y,z)$ and $(a,b,c)$?



I need a hint






inequality cauchy-schwarz-inequality







share|cite|improve this question











$endgroup$







  • 2




    $begingroup$
    This is known as Titus Lemma. More details here
    $endgroup$
    – rtybase
    2 days ago










  • $begingroup$
    Try $(x/sqrt a, y/sqrt b, z/sqrt c)$ and $(sqrt a, sqrt b, sqrt c)$…
    $endgroup$
    – Macavity
    2 days ago













1












1








1





$begingroup$


$a,b,cinBbb R^+, x,y,zin Bbb R, $ show that $fracx^2a+fracy^2b+fracz^2c ge frac(x+y+z)^2a+b+c$ (use Cauchy–Schwarz inequality)



I have trouble finding the two vectors. Is it $(x,y,z)$ and $(a,b,c)$?



I need a hint






inequality cauchy-schwarz-inequality







share|cite|improve this question











$endgroup$




$a,b,cinBbb R^+, x,y,zin Bbb R, $ show that $fracx^2a+fracy^2b+fracz^2c ge frac(x+y+z)^2a+b+c$ (use Cauchy–Schwarz inequality)



I have trouble finding the two vectors. Is it $(x,y,z)$ and $(a,b,c)$?



I need a hint






inequality cauchy-schwarz-inequality




inequality 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

108k1895200




108k1895200










asked 2 days ago









DavidDavid

454




454







  • 2




    $begingroup$
    This is known as Titus Lemma. More details here
    $endgroup$
    – rtybase
    2 days ago










  • $begingroup$
    Try $(x/sqrt a, y/sqrt b, z/sqrt c)$ and $(sqrt a, sqrt b, sqrt c)$…
    $endgroup$
    – Macavity
    2 days ago












  • 2




    $begingroup$
    This is known as Titus Lemma. More details here
    $endgroup$
    – rtybase
    2 days ago










  • $begingroup$
    Try $(x/sqrt a, y/sqrt b, z/sqrt c)$ and $(sqrt a, sqrt b, sqrt c)$…
    $endgroup$
    – Macavity
    2 days ago







2




2




$begingroup$
This is known as Titus Lemma. More details here
$endgroup$
– rtybase
2 days ago




$begingroup$
This is known as Titus Lemma. More details here
$endgroup$
– rtybase
2 days ago












$begingroup$
Try $(x/sqrt a, y/sqrt b, z/sqrt c)$ and $(sqrt a, sqrt b, sqrt c)$…
$endgroup$
– Macavity
2 days ago




$begingroup$
Try $(x/sqrt a, y/sqrt b, z/sqrt c)$ and $(sqrt a, sqrt b, sqrt c)$…
$endgroup$
– Macavity
2 days ago










2 Answers
2






active

oldest

votes


















2












$begingroup$

Hint: Use Cauchy Schwarz in Engelform:
$$fracx^2a+fracy^2b+fracz^2cgeq frac(x+y+z)^2a+b+c$$
It is equivalent to $$z^2a^2b+y^2a^2c+z^2ab^2-2,abcxy-2,abcxz-2,a
bcyz+y^2ac^2+x^2b^2c+x^2bc^2
geq 0$$
and this is $$a(bz-cy)^2+b(az-xc)^2+c(bx-ay)^2geq 0$$ if $$a,b,c$$ are positive.






share|cite|improve this answer











$endgroup$




















    0












    $begingroup$

    By C-S $$fracx^2a+fracy^2b+fracz^2c=fracleft(fracx^2a+fracy^2b+fracz^2cright)(a+b+c)a+b+cgeq$$
    $$geqfracleft(fracxsqrtacdotsqrta+fracysqrtbcdotsqrtb+fraczsqrtccdotsqrtcright)^2a+b+c=frac(x+y+z)^2a+b+c.$$
    The equality occurs for $$left(fracxsqrta,fracysqrtb,fraczsqrtcright)||(sqrta,sqrtb,sqrtc)$$ or
    $$(x,y,z)||(a,b,c).$$
    Actually, the last writing is very useful.






    share|cite|improve this answer









    $endgroup$












      Your Answer





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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2












      $begingroup$

      Hint: Use Cauchy Schwarz in Engelform:
      $$fracx^2a+fracy^2b+fracz^2cgeq frac(x+y+z)^2a+b+c$$
      It is equivalent to $$z^2a^2b+y^2a^2c+z^2ab^2-2,abcxy-2,abcxz-2,a
      bcyz+y^2ac^2+x^2b^2c+x^2bc^2
      geq 0$$
      and this is $$a(bz-cy)^2+b(az-xc)^2+c(bx-ay)^2geq 0$$ if $$a,b,c$$ are positive.






      share|cite|improve this answer











      $endgroup$

















        2












        $begingroup$

        Hint: Use Cauchy Schwarz in Engelform:
        $$fracx^2a+fracy^2b+fracz^2cgeq frac(x+y+z)^2a+b+c$$
        It is equivalent to $$z^2a^2b+y^2a^2c+z^2ab^2-2,abcxy-2,abcxz-2,a
        bcyz+y^2ac^2+x^2b^2c+x^2bc^2
        geq 0$$
        and this is $$a(bz-cy)^2+b(az-xc)^2+c(bx-ay)^2geq 0$$ if $$a,b,c$$ are positive.






        share|cite|improve this answer











        $endgroup$















          2












          2








          2





          $begingroup$

          Hint: Use Cauchy Schwarz in Engelform:
          $$fracx^2a+fracy^2b+fracz^2cgeq frac(x+y+z)^2a+b+c$$
          It is equivalent to $$z^2a^2b+y^2a^2c+z^2ab^2-2,abcxy-2,abcxz-2,a
          bcyz+y^2ac^2+x^2b^2c+x^2bc^2
          geq 0$$
          and this is $$a(bz-cy)^2+b(az-xc)^2+c(bx-ay)^2geq 0$$ if $$a,b,c$$ are positive.






          share|cite|improve this answer











          $endgroup$



          Hint: Use Cauchy Schwarz in Engelform:
          $$fracx^2a+fracy^2b+fracz^2cgeq frac(x+y+z)^2a+b+c$$
          It is equivalent to $$z^2a^2b+y^2a^2c+z^2ab^2-2,abcxy-2,abcxz-2,a
          bcyz+y^2ac^2+x^2b^2c+x^2bc^2
          geq 0$$
          and this is $$a(bz-cy)^2+b(az-xc)^2+c(bx-ay)^2geq 0$$ if $$a,b,c$$ are positive.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 2 days ago

























          answered 2 days ago









          Dr. Sonnhard GraubnerDr. Sonnhard Graubner

          77.7k42866




          77.7k42866





















              0












              $begingroup$

              By C-S $$fracx^2a+fracy^2b+fracz^2c=fracleft(fracx^2a+fracy^2b+fracz^2cright)(a+b+c)a+b+cgeq$$
              $$geqfracleft(fracxsqrtacdotsqrta+fracysqrtbcdotsqrtb+fraczsqrtccdotsqrtcright)^2a+b+c=frac(x+y+z)^2a+b+c.$$
              The equality occurs for $$left(fracxsqrta,fracysqrtb,fraczsqrtcright)||(sqrta,sqrtb,sqrtc)$$ or
              $$(x,y,z)||(a,b,c).$$
              Actually, the last writing is very useful.






              share|cite|improve this answer









              $endgroup$

















                0












                $begingroup$

                By C-S $$fracx^2a+fracy^2b+fracz^2c=fracleft(fracx^2a+fracy^2b+fracz^2cright)(a+b+c)a+b+cgeq$$
                $$geqfracleft(fracxsqrtacdotsqrta+fracysqrtbcdotsqrtb+fraczsqrtccdotsqrtcright)^2a+b+c=frac(x+y+z)^2a+b+c.$$
                The equality occurs for $$left(fracxsqrta,fracysqrtb,fraczsqrtcright)||(sqrta,sqrtb,sqrtc)$$ or
                $$(x,y,z)||(a,b,c).$$
                Actually, the last writing is very useful.






                share|cite|improve this answer









                $endgroup$















                  0












                  0








                  0





                  $begingroup$

                  By C-S $$fracx^2a+fracy^2b+fracz^2c=fracleft(fracx^2a+fracy^2b+fracz^2cright)(a+b+c)a+b+cgeq$$
                  $$geqfracleft(fracxsqrtacdotsqrta+fracysqrtbcdotsqrtb+fraczsqrtccdotsqrtcright)^2a+b+c=frac(x+y+z)^2a+b+c.$$
                  The equality occurs for $$left(fracxsqrta,fracysqrtb,fraczsqrtcright)||(sqrta,sqrtb,sqrtc)$$ or
                  $$(x,y,z)||(a,b,c).$$
                  Actually, the last writing is very useful.






                  share|cite|improve this answer









                  $endgroup$



                  By C-S $$fracx^2a+fracy^2b+fracz^2c=fracleft(fracx^2a+fracy^2b+fracz^2cright)(a+b+c)a+b+cgeq$$
                  $$geqfracleft(fracxsqrtacdotsqrta+fracysqrtbcdotsqrtb+fraczsqrtccdotsqrtcright)^2a+b+c=frac(x+y+z)^2a+b+c.$$
                  The equality occurs for $$left(fracxsqrta,fracysqrtb,fraczsqrtcright)||(sqrta,sqrtb,sqrtc)$$ or
                  $$(x,y,z)||(a,b,c).$$
                  Actually, the last writing is very useful.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 2 days ago









                  Michael RozenbergMichael Rozenberg

                  108k1895200




                  108k1895200



























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