Let $a,b,c>0$,and $a+b+c=3$,show that $$\dfrac{a}{2b^3+c}+\dfrac{b}{2c^3+a}+\dfrac{c}{2a^3+b}\ge 1$$

such Use Cauchy-Schwarz inequality we have $$\left(\dfrac{a}{2b^3+c}+\dfrac{b}{2c^3+a}+\dfrac{c}{2a^3+b}\right)\left(a(2b^3+c)+b(2c^3+a)+c(2a^3+b)\right)\ge (a+b+c)^2=9$$

Therefore,it suffices to prove that $$(2ab^3+2bc^3+2ca^3)+(ab+bc+ca)\le 9$$ The last inequality doesn't hold for $a=1,b=1.9$,then $2ab^3>9$ I just do it now

Thanks in advance!


By C-S $\sum\limits_{cyc}\frac{a}{2b^3+c}=\sum\limits_{cyc}\frac{a^2(a+c)^2}{a(a+c)^2(2b^3+c)}\geq\frac{\left(\sum\limits_{cyc}(a^2+ab)\right)^2}{\sum\limits_{cyc}a(a+c)^2(2b^3+c)}$.

Hence, it remains to prove that $(a+b+c)^2\left(\sum\limits_{cyc}(a^2+ab)\right)^2\geq9\sum\limits_{cyc}a(a+c)^2(2b^3+c)$, which is

$\sum\limits_{cyc}(a^6+3a^5b+3a^5c+4a^4b^2+4a^4c^2-14a^3b^3+10a^4bc-a^3b^2c-19a^3c^2b+9a^2b^2c^2)\geq0$, which is obvious.

For example, $LS\geq\sum\limits_{cyc}(a^6-a^5b-a^5c+a^4bc)+\sum\limits_{cyc}(3a^5b+3a^5c+4a^4b^2+4a^4c^2-14a^3b^3)+$


  • $\begingroup$ Nice! Thank you very much $\endgroup$ – partofsha May 5 '16 at 12:22
  • $\begingroup$ maybe $\sum\dfrac{a}{nb^n+c}\ge \dfrac{3}{n+1}?$ $\endgroup$ – partofsha May 5 '16 at 12:23
  • $\begingroup$ Sorry,but I want to ask where is the term $9a^3c$ $\endgroup$ – cxz May 7 '16 at 1:58
  • $\begingroup$ @cxz $9c=(a+b+c)^2c$. $\endgroup$ – Michael Rozenberg May 7 '16 at 2:17
  • $\begingroup$ @MichaelRozenberg yeah,i got it now, but how did you came up with this,i mean, even if you got the long formula, i don't it's that obvious, one can't tell if it is true at first glance $\endgroup$ – cxz May 7 '16 at 2:22

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