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 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 $a_1=a_2=\ldots=a_n=1$.
