How to prove inequality $(2n-1)x^{-1} + (2n+1)x^{2n-1} > \frac{(2n+1)^2}{2n}, \quad n > 2$ and $ 0 < x < 1$

Question

How to prove the following inequality for $n > 2$ and $ 0 < x < 1$:

$$(2n-1)x^{-1} + (2n+1)x^{2n-1} > \frac{(2n+1)^2}{2n}, \quad n \in \mathbb{N}, x \in \mathbb{R}$$

I tried using weighted arithmetic mean - geometric mean inequality, but then I end up with:

$$2n \sqrt[2n]{2n+1} \ge \frac{(2n + 1)^2}{2n}$$

which don't work.

Answer 1

Now, we need to prove that $$(2n+1)^{\frac{1}{2n}}>\left(\frac{2n+1}{2n}\right)^2$$ or $$\sqrt{2n+1}>\left(1+\frac{1}{2n}\right)^{2n},$$ which is obvious for any $n\geq4$.

Can you end it now?