Prove that $frac{a}{csqrt{a^2+1}}+frac{b}{asqrt{b^2+1}}+frac{c}{bsqrt{c^2+1}}ge frac{3}{2}$












1












$begingroup$


Let $a,b,cin Bbb R^+$ such that $a+b+c=abc$. Prove that $$frac{a}{csqrt{a^2+1}}+frac{b}{asqrt{b^2+1}}+frac{c}{bsqrt{c^2+1}}ge frac{3}{2}$$




Idea 1.From $a+b+c=abcLeftrightarrow frac{1}{ab}+frac{1}{bc}+frac{1}{ca}=1$. Let $left(frac{1}{a};frac{1}{b};frac{1}{c}right)rightarrow left(x;y;zright)$



So i need to prove $frac{z}{sqrt{1+x^2}}+frac{y}{sqrt{z^2+1}}+frac{x}{sqrt{y^2+1}}ge frac{3}{2}$



By AM-GM $frac{x}{sqrt{y^2+1}}=frac{x}{sqrt{left(x+yright)left(y+zright)}}ge frac{2x}{x+2y+z}$



$$LHSge 2sum _{cyc}frac{x}{x+2y+z}=2sum _{cyc}frac{x^2}{x^2+2xy+xz}ge 2frac{left(x+y+zright)^2}{sum _{cyc}x^2+sum _{cyc}3xy}$$



Or $4left(x+y+zright)^2ge 3left(x^2+y^2+z^2+3xy+3yz+3xzright)$



Or $x^2+y^2+z^2ge xy+yz+xz$ (true)



Idea 2. By Holder $$left(sum _{cyc}frac{a}{csqrt{a^2+1}}right)left(sum _{cyc}frac{a}{csqrt{a^2+1}}right)sum _{cyc}left(c^2left(a^2+1right)aright)ge left(sum _{cyc}aright)^3$$



I will prove the inequality $frac{left(a+b+cright)^3}{c^2aleft(a^2+1right)+a^2bleft(b^2+1right)+b^2cleft(c^2+1right)}ge frac{3}{2}$



Or $frac{abcleft(a+b+cright)^3}{left(a+b+cright)left(a^2b^3+b^2c^3+c^2a^3right)+abcleft(a^2b+b^2c+c^2aright)}ge frac{3}{2}$



I tried $SOS$ but failed help me improve "idea 2" use Holder






inequality radicals geometric-inequalities holder-inequality rearrangement-inequality







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    Let $a,b,cin Bbb R^+$ such that $a+b+c=abc$. Prove that $$frac{a}{csqrt{a^2+1}}+frac{b}{asqrt{b^2+1}}+frac{c}{bsqrt{c^2+1}}ge frac{3}{2}$$




    Idea 1.From $a+b+c=abcLeftrightarrow frac{1}{ab}+frac{1}{bc}+frac{1}{ca}=1$. Let $left(frac{1}{a};frac{1}{b};frac{1}{c}right)rightarrow left(x;y;zright)$



    So i need to prove $frac{z}{sqrt{1+x^2}}+frac{y}{sqrt{z^2+1}}+frac{x}{sqrt{y^2+1}}ge frac{3}{2}$



    By AM-GM $frac{x}{sqrt{y^2+1}}=frac{x}{sqrt{left(x+yright)left(y+zright)}}ge frac{2x}{x+2y+z}$



    $$LHSge 2sum _{cyc}frac{x}{x+2y+z}=2sum _{cyc}frac{x^2}{x^2+2xy+xz}ge 2frac{left(x+y+zright)^2}{sum _{cyc}x^2+sum _{cyc}3xy}$$



    Or $4left(x+y+zright)^2ge 3left(x^2+y^2+z^2+3xy+3yz+3xzright)$



    Or $x^2+y^2+z^2ge xy+yz+xz$ (true)



    Idea 2. By Holder $$left(sum _{cyc}frac{a}{csqrt{a^2+1}}right)left(sum _{cyc}frac{a}{csqrt{a^2+1}}right)sum _{cyc}left(c^2left(a^2+1right)aright)ge left(sum _{cyc}aright)^3$$



    I will prove the inequality $frac{left(a+b+cright)^3}{c^2aleft(a^2+1right)+a^2bleft(b^2+1right)+b^2cleft(c^2+1right)}ge frac{3}{2}$



    Or $frac{abcleft(a+b+cright)^3}{left(a+b+cright)left(a^2b^3+b^2c^3+c^2a^3right)+abcleft(a^2b+b^2c+c^2aright)}ge frac{3}{2}$



    I tried $SOS$ but failed help me improve "idea 2" use Holder






    inequality radicals geometric-inequalities holder-inequality rearrangement-inequality







    share|cite|improve this question











    $endgroup$















      1












      1








      1


      1



      $begingroup$


      Let $a,b,cin Bbb R^+$ such that $a+b+c=abc$. Prove that $$frac{a}{csqrt{a^2+1}}+frac{b}{asqrt{b^2+1}}+frac{c}{bsqrt{c^2+1}}ge frac{3}{2}$$




      Idea 1.From $a+b+c=abcLeftrightarrow frac{1}{ab}+frac{1}{bc}+frac{1}{ca}=1$. Let $left(frac{1}{a};frac{1}{b};frac{1}{c}right)rightarrow left(x;y;zright)$



      So i need to prove $frac{z}{sqrt{1+x^2}}+frac{y}{sqrt{z^2+1}}+frac{x}{sqrt{y^2+1}}ge frac{3}{2}$



      By AM-GM $frac{x}{sqrt{y^2+1}}=frac{x}{sqrt{left(x+yright)left(y+zright)}}ge frac{2x}{x+2y+z}$



      $$LHSge 2sum _{cyc}frac{x}{x+2y+z}=2sum _{cyc}frac{x^2}{x^2+2xy+xz}ge 2frac{left(x+y+zright)^2}{sum _{cyc}x^2+sum _{cyc}3xy}$$



      Or $4left(x+y+zright)^2ge 3left(x^2+y^2+z^2+3xy+3yz+3xzright)$



      Or $x^2+y^2+z^2ge xy+yz+xz$ (true)



      Idea 2. By Holder $$left(sum _{cyc}frac{a}{csqrt{a^2+1}}right)left(sum _{cyc}frac{a}{csqrt{a^2+1}}right)sum _{cyc}left(c^2left(a^2+1right)aright)ge left(sum _{cyc}aright)^3$$



      I will prove the inequality $frac{left(a+b+cright)^3}{c^2aleft(a^2+1right)+a^2bleft(b^2+1right)+b^2cleft(c^2+1right)}ge frac{3}{2}$



      Or $frac{abcleft(a+b+cright)^3}{left(a+b+cright)left(a^2b^3+b^2c^3+c^2a^3right)+abcleft(a^2b+b^2c+c^2aright)}ge frac{3}{2}$



      I tried $SOS$ but failed help me improve "idea 2" use Holder






      inequality radicals geometric-inequalities holder-inequality rearrangement-inequality







      share|cite|improve this question











      $endgroup$




      Let $a,b,cin Bbb R^+$ such that $a+b+c=abc$. Prove that $$frac{a}{csqrt{a^2+1}}+frac{b}{asqrt{b^2+1}}+frac{c}{bsqrt{c^2+1}}ge frac{3}{2}$$




      Idea 1.From $a+b+c=abcLeftrightarrow frac{1}{ab}+frac{1}{bc}+frac{1}{ca}=1$. Let $left(frac{1}{a};frac{1}{b};frac{1}{c}right)rightarrow left(x;y;zright)$



      So i need to prove $frac{z}{sqrt{1+x^2}}+frac{y}{sqrt{z^2+1}}+frac{x}{sqrt{y^2+1}}ge frac{3}{2}$



      By AM-GM $frac{x}{sqrt{y^2+1}}=frac{x}{sqrt{left(x+yright)left(y+zright)}}ge frac{2x}{x+2y+z}$



      $$LHSge 2sum _{cyc}frac{x}{x+2y+z}=2sum _{cyc}frac{x^2}{x^2+2xy+xz}ge 2frac{left(x+y+zright)^2}{sum _{cyc}x^2+sum _{cyc}3xy}$$



      Or $4left(x+y+zright)^2ge 3left(x^2+y^2+z^2+3xy+3yz+3xzright)$



      Or $x^2+y^2+z^2ge xy+yz+xz$ (true)



      Idea 2. By Holder $$left(sum _{cyc}frac{a}{csqrt{a^2+1}}right)left(sum _{cyc}frac{a}{csqrt{a^2+1}}right)sum _{cyc}left(c^2left(a^2+1right)aright)ge left(sum _{cyc}aright)^3$$



      I will prove the inequality $frac{left(a+b+cright)^3}{c^2aleft(a^2+1right)+a^2bleft(b^2+1right)+b^2cleft(c^2+1right)}ge frac{3}{2}$



      Or $frac{abcleft(a+b+cright)^3}{left(a+b+cright)left(a^2b^3+b^2c^3+c^2a^3right)+abcleft(a^2b+b^2c+c^2aright)}ge frac{3}{2}$



      I tried $SOS$ but failed help me improve "idea 2" use Holder






      inequality radicals geometric-inequalities holder-inequality rearrangement-inequality




      inequality radicals geometric-inequalities holder-inequality rearrangement-inequality






      share|cite|improve this question















      share|cite|improve this question













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      edited Jan 8 at 19:09









      Michael Rozenberg

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      asked Jan 8 at 15:07









      Nguyễn Duy LinhNguyễn Duy Linh

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          $begingroup$

          We need to prove that
          $$sum_{cyc}frac{a}{csqrt{a^2+frac{abc}{a+b+c}}}geqfrac{3}{2}$$ or
          $$sum_{cyc}sqrt{frac{a(a+b+c)}{c^2(a+b)(a+c)}}geqfrac{3}{2}$$ or
          $$sum_{cyc}sqrt{frac{a^2b}{c(a+b)(a+c)}}geqfrac{3}{2}$$ or
          $$sum_{cyc }sqrt{a^3b^2(b+c)}geqfrac{3}{2}sqrt{abc(a+b)(a+c)(b+c)}$$ or
          $$sum_{cyc }sqrt{(a^3b^3+a^3b^2c)}geqfrac{3}{2}sqrt{abc(a+b)(a+c)(b+c)}.$$
          Now, let $ab=z$, $ac=y$ and $bc=x$.



          Thus, we need to prove that
          $$sum_{cyc}sqrt{z^3+z^2x}geqfrac{3}{2}sqrt{(x+y)(x+z)(y+z)}.$$
          By Holder
          $$left(sum_{cyc}sqrt{z^3+z^2x}right)^2sum_{cyc}frac{z}{z+x}geq(x+y+z)^3.$$
          Id est, it's enough to prove that
          $$4(x+y+z)^3geq9prod_{cyc}(x+y)sum_{cyc}frac{z}{z+x}$$ or
          $$4(x+y+z)^3geq9sum_{cyc}x(x+z)(y+z)$$ or
          $$4(x+y+z)^3geq9sum_{cyc}(2x^2y+x^2z+xyz).$$
          Now, by Rearrangement easy to show that $$x^2y+y^2z+z^2x+xyzleqfrac{4}{27}(x+y+z)^3.$$
          Thus, it's enough to prove that
          $$4(x+y+z)^3geqfrac{4}{3}(x+y+z)^3+9sum_{cyc}left(x^2y+x^2z+frac{2}{3}xyzright)$$ or
          $$frac{8}{3}(x+y+z)^3geq9(x+y)(x+z)(y+z)$$ or
          $$left(frac{x+y+x+z+y+z}{3}right)^3geq(x+y)(x+z)(y+z),$$ which is true by AM-GM.



          Done!






          share|cite|improve this answer











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            $begingroup$

            We need to prove that
            $$sum_{cyc}frac{a}{csqrt{a^2+frac{abc}{a+b+c}}}geqfrac{3}{2}$$ or
            $$sum_{cyc}sqrt{frac{a(a+b+c)}{c^2(a+b)(a+c)}}geqfrac{3}{2}$$ or
            $$sum_{cyc}sqrt{frac{a^2b}{c(a+b)(a+c)}}geqfrac{3}{2}$$ or
            $$sum_{cyc }sqrt{a^3b^2(b+c)}geqfrac{3}{2}sqrt{abc(a+b)(a+c)(b+c)}$$ or
            $$sum_{cyc }sqrt{(a^3b^3+a^3b^2c)}geqfrac{3}{2}sqrt{abc(a+b)(a+c)(b+c)}.$$
            Now, let $ab=z$, $ac=y$ and $bc=x$.



            Thus, we need to prove that
            $$sum_{cyc}sqrt{z^3+z^2x}geqfrac{3}{2}sqrt{(x+y)(x+z)(y+z)}.$$
            By Holder
            $$left(sum_{cyc}sqrt{z^3+z^2x}right)^2sum_{cyc}frac{z}{z+x}geq(x+y+z)^3.$$
            Id est, it's enough to prove that
            $$4(x+y+z)^3geq9prod_{cyc}(x+y)sum_{cyc}frac{z}{z+x}$$ or
            $$4(x+y+z)^3geq9sum_{cyc}x(x+z)(y+z)$$ or
            $$4(x+y+z)^3geq9sum_{cyc}(2x^2y+x^2z+xyz).$$
            Now, by Rearrangement easy to show that $$x^2y+y^2z+z^2x+xyzleqfrac{4}{27}(x+y+z)^3.$$
            Thus, it's enough to prove that
            $$4(x+y+z)^3geqfrac{4}{3}(x+y+z)^3+9sum_{cyc}left(x^2y+x^2z+frac{2}{3}xyzright)$$ or
            $$frac{8}{3}(x+y+z)^3geq9(x+y)(x+z)(y+z)$$ or
            $$left(frac{x+y+x+z+y+z}{3}right)^3geq(x+y)(x+z)(y+z),$$ which is true by AM-GM.



            Done!






            share|cite|improve this answer











            $endgroup$


















              2












              $begingroup$

              We need to prove that
              $$sum_{cyc}frac{a}{csqrt{a^2+frac{abc}{a+b+c}}}geqfrac{3}{2}$$ or
              $$sum_{cyc}sqrt{frac{a(a+b+c)}{c^2(a+b)(a+c)}}geqfrac{3}{2}$$ or
              $$sum_{cyc}sqrt{frac{a^2b}{c(a+b)(a+c)}}geqfrac{3}{2}$$ or
              $$sum_{cyc }sqrt{a^3b^2(b+c)}geqfrac{3}{2}sqrt{abc(a+b)(a+c)(b+c)}$$ or
              $$sum_{cyc }sqrt{(a^3b^3+a^3b^2c)}geqfrac{3}{2}sqrt{abc(a+b)(a+c)(b+c)}.$$
              Now, let $ab=z$, $ac=y$ and $bc=x$.



              Thus, we need to prove that
              $$sum_{cyc}sqrt{z^3+z^2x}geqfrac{3}{2}sqrt{(x+y)(x+z)(y+z)}.$$
              By Holder
              $$left(sum_{cyc}sqrt{z^3+z^2x}right)^2sum_{cyc}frac{z}{z+x}geq(x+y+z)^3.$$
              Id est, it's enough to prove that
              $$4(x+y+z)^3geq9prod_{cyc}(x+y)sum_{cyc}frac{z}{z+x}$$ or
              $$4(x+y+z)^3geq9sum_{cyc}x(x+z)(y+z)$$ or
              $$4(x+y+z)^3geq9sum_{cyc}(2x^2y+x^2z+xyz).$$
              Now, by Rearrangement easy to show that $$x^2y+y^2z+z^2x+xyzleqfrac{4}{27}(x+y+z)^3.$$
              Thus, it's enough to prove that
              $$4(x+y+z)^3geqfrac{4}{3}(x+y+z)^3+9sum_{cyc}left(x^2y+x^2z+frac{2}{3}xyzright)$$ or
              $$frac{8}{3}(x+y+z)^3geq9(x+y)(x+z)(y+z)$$ or
              $$left(frac{x+y+x+z+y+z}{3}right)^3geq(x+y)(x+z)(y+z),$$ which is true by AM-GM.



              Done!






              share|cite|improve this answer











              $endgroup$
















                2












                2








                2





                $begingroup$

                We need to prove that
                $$sum_{cyc}frac{a}{csqrt{a^2+frac{abc}{a+b+c}}}geqfrac{3}{2}$$ or
                $$sum_{cyc}sqrt{frac{a(a+b+c)}{c^2(a+b)(a+c)}}geqfrac{3}{2}$$ or
                $$sum_{cyc}sqrt{frac{a^2b}{c(a+b)(a+c)}}geqfrac{3}{2}$$ or
                $$sum_{cyc }sqrt{a^3b^2(b+c)}geqfrac{3}{2}sqrt{abc(a+b)(a+c)(b+c)}$$ or
                $$sum_{cyc }sqrt{(a^3b^3+a^3b^2c)}geqfrac{3}{2}sqrt{abc(a+b)(a+c)(b+c)}.$$
                Now, let $ab=z$, $ac=y$ and $bc=x$.



                Thus, we need to prove that
                $$sum_{cyc}sqrt{z^3+z^2x}geqfrac{3}{2}sqrt{(x+y)(x+z)(y+z)}.$$
                By Holder
                $$left(sum_{cyc}sqrt{z^3+z^2x}right)^2sum_{cyc}frac{z}{z+x}geq(x+y+z)^3.$$
                Id est, it's enough to prove that
                $$4(x+y+z)^3geq9prod_{cyc}(x+y)sum_{cyc}frac{z}{z+x}$$ or
                $$4(x+y+z)^3geq9sum_{cyc}x(x+z)(y+z)$$ or
                $$4(x+y+z)^3geq9sum_{cyc}(2x^2y+x^2z+xyz).$$
                Now, by Rearrangement easy to show that $$x^2y+y^2z+z^2x+xyzleqfrac{4}{27}(x+y+z)^3.$$
                Thus, it's enough to prove that
                $$4(x+y+z)^3geqfrac{4}{3}(x+y+z)^3+9sum_{cyc}left(x^2y+x^2z+frac{2}{3}xyzright)$$ or
                $$frac{8}{3}(x+y+z)^3geq9(x+y)(x+z)(y+z)$$ or
                $$left(frac{x+y+x+z+y+z}{3}right)^3geq(x+y)(x+z)(y+z),$$ which is true by AM-GM.



                Done!






                share|cite|improve this answer











                $endgroup$



                We need to prove that
                $$sum_{cyc}frac{a}{csqrt{a^2+frac{abc}{a+b+c}}}geqfrac{3}{2}$$ or
                $$sum_{cyc}sqrt{frac{a(a+b+c)}{c^2(a+b)(a+c)}}geqfrac{3}{2}$$ or
                $$sum_{cyc}sqrt{frac{a^2b}{c(a+b)(a+c)}}geqfrac{3}{2}$$ or
                $$sum_{cyc }sqrt{a^3b^2(b+c)}geqfrac{3}{2}sqrt{abc(a+b)(a+c)(b+c)}$$ or
                $$sum_{cyc }sqrt{(a^3b^3+a^3b^2c)}geqfrac{3}{2}sqrt{abc(a+b)(a+c)(b+c)}.$$
                Now, let $ab=z$, $ac=y$ and $bc=x$.



                Thus, we need to prove that
                $$sum_{cyc}sqrt{z^3+z^2x}geqfrac{3}{2}sqrt{(x+y)(x+z)(y+z)}.$$
                By Holder
                $$left(sum_{cyc}sqrt{z^3+z^2x}right)^2sum_{cyc}frac{z}{z+x}geq(x+y+z)^3.$$
                Id est, it's enough to prove that
                $$4(x+y+z)^3geq9prod_{cyc}(x+y)sum_{cyc}frac{z}{z+x}$$ or
                $$4(x+y+z)^3geq9sum_{cyc}x(x+z)(y+z)$$ or
                $$4(x+y+z)^3geq9sum_{cyc}(2x^2y+x^2z+xyz).$$
                Now, by Rearrangement easy to show that $$x^2y+y^2z+z^2x+xyzleqfrac{4}{27}(x+y+z)^3.$$
                Thus, it's enough to prove that
                $$4(x+y+z)^3geqfrac{4}{3}(x+y+z)^3+9sum_{cyc}left(x^2y+x^2z+frac{2}{3}xyzright)$$ or
                $$frac{8}{3}(x+y+z)^3geq9(x+y)(x+z)(y+z)$$ or
                $$left(frac{x+y+x+z+y+z}{3}right)^3geq(x+y)(x+z)(y+z),$$ which is true by AM-GM.



                Done!







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Jan 8 at 19:06

























                answered Jan 8 at 19:01









                Michael RozenbergMichael Rozenberg

                100k1591193




                100k1591193






























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                    A Topological Invariant for $pi_3(U(n))$

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                    3 3 $begingroup$ Recently, I saw a construction of topological invariant for $pi_3(U(n))$ with $ngeq 2$ : $$ N=frac{1}{24pi^2}int_{S^3} d^3x epsilon^{ijk} Tr[(U^{-1}partial_{x_i}U)(U^{-1}partial_{x_j}U)(U^{-1}partial_{x_k}U)] , $$ where $Uin U(n)$ depends on $boldsymbol{x}=(x_1,x_2,x_3)in S^3$ , $epsilon^{ijk}$ is the Levi-Civita symbol, $i,j,k=1,2,3$ , and the duplicated indexes are summed over. It is claimed that $N$ is an integer, but why? Update 02/02/2019 I think I got an argument for $n=2$ . In this case, $U=e^{i varphi} q$ with $qin SU(2)$ . Due to the trace and the Levi-Civita symbol in $N$ , $varphi$ does not contribute to $N$ . As $Tr[q^{dagger}partial_i q]=0$ and $(q^{dagger}partial_i q)^{dagger}=-q^{dagger}partial_i q$ , $q^{dagger}partial_i q$ in geneeral has the form $q^{dagger}partia
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