Typical Olympiad Inequality? If $\sum_i^na_i=n$ with $a_i>0$, then $\sum_{i=1}^n\left(\frac{a_i^3+1}{a_i^2+1}\right)^4\geq n$

You need to use another queue:

By Rearrangement, AM-GM and C-S we obtain: $$\sum_{i=1}^n\left(\frac{a_i^3+1}{a_i^2+1}\right)^4\geq \ \sum_{i=1}^n\left(\frac{a_i+1}{2}\right)^4\geq\sum_{i=1}^na_i^2=\frac{1}{n}\sum_{i=1}^n1^2\sum_{i=1}^na_i^2\geq\frac{1}{n}\left(\sum_{i=1}^na_i\right)^2=n. $$

I used the following. $$\frac{x^3+1}{x^2+1}\geq\frac{x+1}{2}$$ it's $$2(x^3+1)\geq(x^2+1)(x+1)$$ or $$x^3+1\geq x^2+x$$ or $$x^2\cdot x+1\cdot1\geq x^2\cdot1+x\cdot1,$$ which is true by Rearrangement because $(x^2,1)$ and $(x,1)$ have the same ordering.

Also, by AM-GM $$\left(\frac{x+1}{2}\right)^4\geq\left(\sqrt{x}\right)^4=x^2.$$

A Hint for an Alternative Solution.

We want to show that $$\left(\frac{x^3+1}{x^2+1}\right)^4\geq 2x-1$$ for every $x\in\mathbb{R}$. By the AM-GM Inequality, $$\left(\frac{x^3+1}{x^2+1}\right)^4+1\geq 2\,\left(\frac{x^3+1}{x^2+1}\right)^2\,.$$ Hence, it suffices to verify that $$\left(\frac{x^3+1}{x^2+1}\right)^2\geq x$$ for all $x\in\mathbb{R}$. This part is left for the OP.

Remark. From this solution, we need not require that $a_1,a_2,\ldots,a_n$ be positive. That is, for any real numbers $a_1,a_2,\ldots,a_n$ such that $\sum\limits_{i=1}^n\,a_i=n$, we always have $$\sum_{i=1}^n\,\left(\frac{a_i^3+1}{a_i^2+1}\right)^4\geq n\,.$$ However, the sole equality case is when $a_1=a_2=\ldots=a_n=1$.