How to prove the following inequality with two variables How to prove the following inequality, which should hold for all $x \in \left]0; 1\right[$, $n \in \mathbb{N}$, n is even.
\begin{align}
n^2 x^{n-2} (x^{n+1}+1)^3\left[2nx^n - (n-1)(x^n-1)\right] > (n+1)^2 x^{n+1} (x^n-1)^3\left[(2(n+1)x^{n+1} - n(x^{n+1}+1)\right]
\end{align}
My attempt so far:
\begin{align}
n^2 x^{n-2} (x^{n+1}+1)^3\left[2nx^n - (n-1)(x^n-1)\right] &> (n+1)^2 x^{n+1} (x^n-1)^3\left[2(n+1)x^{n+1} - n(x^{n+1}+1)\right] \\
n^2 (x^{n+1}+1)^3\left[2nx^n - (n-1)(x^n-1)\right] &> (n+1)^2 x^3 (x^n-1)^3\left[2(n+1)x^{n+1} - n(x^{n+1}+1)\right] \\
\left(\frac{n}{n+1}\right)^2 \left(\frac{x^{n+1}+1}{x^{n+3}-1}\right)^3\frac{2nx^n - (n-1)(x^n-1)}{2(n+1)x^{n+1} - n(x^{n+1}+1)} &> 1
\end{align}
\begin{align}
\frac{1}{4} &\leq \left(\frac{n}{n+1}\right)^2 &< 1 \\
-1 &< \left(\frac{x^{n+1}+1}{x^{n+3}-1}\right)^3 &< 0
\end{align}
But here I stuck.

Indeed you need to prove:
\begin{align}
\left(\frac{n}{n+1}\right)^2 \left(\frac{1+x^{n+1}}{x-x^{n+1}}\right)^3\frac{(n-1)+(n+1)x^n}{n-(n+2)x^{n+1}} &> 1
\end{align}
We have
$$ |n-(n+2)x^{n+1}|\leq n.$$
Also
$$ \left(\frac{1+x^{n+1}}{x-x^{n+1}}\right)^3>x^{-3},$$
so we need to prove that
$$(n-1)x^{-3}+(n+1)x^{n-3}>\frac{(n+1)^2}{n}.$$
For $n\geq 4$ we use Cauchy in the next way:
$$(n-1)x^{-3}+(n+1)x^{n-3}=(n-3)*\frac{n-1}{n-3}x^{-3}+3*\frac{n+1}{3}x^{n-3}\geq$$
$$\geq n \sqrt[n]{\left (\frac{n-1}{n-3} \right )^{n-3} \left (\frac{n+1}{3} \right )^{3}}\geq n*\min \left (\frac{n-1}{n-3},\frac{n+1}{3}\right ).$$
The last expression is more than $(n+1)^2/n$ if $n\geq 4$.
If $n=2$ we need to prove:
$$(1+x^3)^3(1+3x^2)>\frac{9}{2}x^3(1-x^2)^3(1-2x^3),$$
which is true because the right side is always less than $1$ while left side is more than $1$.
