Inequality : $\sum_{cyc}\left(\frac{a^3+1}{a^2+1}\right)^4 \geq \frac{1}{27}\left(\sum_{cyc}\sqrt{ab}\right)^4$

Question

Let $\{a, b, c\} \subset \mathbb{R}^+$, where $a+b+c=3$. Prove that: $$\left(\frac{a^3+1}{a^2+1}\right)^4 + \left(\frac{b^3+1}{b^2+1}\right)^4 + \left(\frac{c^3+1}{c^2+1}\right)^4 \geq \frac{1}{27}\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right)^4.$$

My attempt :

$a+b+c=3$, so $3 \geq \sqrt{ab}+\sqrt{bc}+\sqrt{ca}$

By Holder Inequality,

$\left(\displaystyle\sum_{cyc}\left(\frac{a^3+1}{a^2+1}\right)^4\right)(\displaystyle\sum_{cyc}1)(\displaystyle\sum_{cyc}1) (\displaystyle\sum_{cyc}1) \geq \left(\displaystyle\sum_{cyc}\frac{a^3+1}{a^2+1}\right)^4$

$27\displaystyle\sum_{cyc}\left(\frac{a^3+1}{a^2+1}\right)^4 \geq \left(\displaystyle\sum_{cyc}\frac{a^3+1}{a^2+1}\right)^4$

We have to show that

$\displaystyle\sum_{cyc}\frac{a^3+1}{a^2+1}\geq \displaystyle\sum_{cyc}\sqrt{ab}$

Answer 1

Since $$ab+ac+bc\leq\frac{1}{3}(a+b+c)^2=3,$$ it's enough to prove that $$\sum_{cyc}\frac{a^3+1}{a^2+1}\geq3$$ or $$\sum_{cyc}\left(\frac{a^3+1}{a^2+1}-1\right)\geq0$$ or $$\sum_{cyc}\frac{a^2(a-1)}{a^2+1}\geq0$$ or $$\sum_{cyc}\left(\frac{a^2(a-1)}{a^2+1}-\frac{1}{2}(a-1)\right)\geq0$$ or $$\sum_{cyc}\frac{(a-1)^2(a+1)}{a^2+1}\geq0.$$ Done!