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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 $a$,$b$,$c$. $$(\sum \limits_{cyc}(a^2+a+1))(\sum \limits_{cyc}(a^6+a^3+1))\geq (\sum \limits_{cyc}(a^4+a^2+1))^2$$

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user2838619
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  • As an aside, you could prove that inequality by using CS: $ ( 1 + 1 + 1 ) ( a^4 + a^2 + 1) \geq ( a^2 + a + 1 ) ^2$ – Calvin Lin Mar 11 '23 at 03:32

1 Answers1

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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.

Jaehyeon Seo
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