# Prove ${\sum \limits_{cyc}\frac{a^4+a^2+1}{a^6+a^3+1}\leq\sum \limits_{cyc}\frac{3}{a^2+a+1}}$

If $a,b,c$ are positive real numbers,Prove:$${\sum \limits_{cyc}\frac{a^4+a^2+1}{a^6+a^3+1}\leq\sum \limits_{cyc}\frac{3}{a^2+a+1}}$$

Additional info: We should only use Cauchy and AM-GM.

Things I have done so far: For $a$ I can write using Cauchy inequality that$$(a^2+a+1)(a^6+a^3+1)\geq (a^4+a^2+1)^2$$$$\frac{a^2+a+1}{a^4+a^2+1}\geq \frac{a^4+a^2+1}{a^6+a^3+1}$$ So i should Prove this: $$\frac{3}{a^2+a+1}\geq\frac{a^2+a+1}{a^4+a^2+1}$$

Working on this inequality lead me to this$$(a-1)^2(a^2+a+1)\geq0$$

which is obviously true.So i can do the same thing for $b$ and $c$ to prove inequality.however I think $\frac{3}{a^2+a+1}\geq\frac{a^2+a+1}{a^4+a^2+1}$ could have counterexample(I'm trying to find it).

And using $\sum \limits_{cyc}$ confuses me and I think below inequality is not true.So this is the reason I do it separately for $a$,$b$,$c$. $$(\sum \limits_{cyc}(a^2+a+1))(\sum \limits_{cyc}(a^6+a^3+1))\geq (\sum \limits_{cyc}(a^4+a^2+1))^2$$

I don't think your proof have a problem. Can I ask for the reason why you are looking for a counterexample?

And for the second one: \begin{align} \left(\sum \limits_{cyc}(a^2+a+1)\right)\left(\sum \limits_{cyc}(a^6+a^3+1)\right) &\ge \left(\sum_{cyc}\left((a^2+a+1)(a^6+a^3+1)\right)^{1/2}\right)^2 \\&\ge \left(\sum_{cyc}(a^4+a^2+1)\right)^2, \end{align} so it holds. But it doesn't seem to help proving the problem.

• well, i ran in to problem in past here,my first inequality did not hold there.so it is for bad memories. – user2838619 Aug 7 '14 at 11:59
• @user2838619 In this case, the term in the summation only depends on one variable, so seperating the inequality may helps. – The Great Seo Aug 7 '14 at 12:00