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

I have tried AM-GM, Cauchy-Schwarz, Rearrangement etc. but nothing seems to work. The fourth power in the LHS really evades me, and I struggle to see what can be done.
My attempts didn’t lead me to any result ... Simply cauchy , where $a_i=x,$ $b_i=1$ to find an inequality involving $\sum x^2$ . I also tried finding an inequality involving $\sum x^3$ using $a_i=\frac{x^3}{2}$ and $b_i=x^{\frac{1}{2}}$
 A: 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: 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 $a_1=a_2=\ldots=a_n=1$.
