3
$\begingroup$

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!

$\endgroup$
3
$\begingroup$

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

$+abc\sum\limits_{cyc}(11a^3-a^2b-19a^2c+9abc)\geq0$.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.