It's related to my own question Refinement inequality of : $\sqrt{x}+x^{\frac{x}{x+1}}\geq x+1$
Let $x\geq 5$ be a real number then we have : $$\frac{x^2+1}{x+1}\Bigg(\frac{x^{\frac{x}{x+1}}}{x^{\frac{x^2}{x^2+1}}}+\frac{\sqrt{x}}{x^{\frac{x}{x+1}}}\Bigg)\geq x+1$$
Now the problem is:
Let :$$\frac{x^{\alpha}+1}{x+1}\Bigg(\frac{x^{\frac{x}{x+1}}}{x^{\frac{x^{\alpha}}{x^{\alpha}+1}}}+\frac{\sqrt{x}}{x^{\frac{x}{x+1}}}\Bigg)\geq x+1$$
Prove that $\alpha=2$ is the best constant, i.e. the above holds for $\alpha=2$ and fails for $\alpha<2$.
I have tried to prove $0<\epsilon<2$:
$$\lim_{x\to \infty}\frac{x^{2-\epsilon}+1}{x+1}\Bigg(\frac{x^{\frac{x}{x+1}}}{x^{\frac{x^{2-\epsilon}}{x^{2-\epsilon}+1}}}+\frac{\sqrt{x}}{x^{\frac{x}{x+1}}}\Bigg) -(x+1)=-\infty$$
Using the series provided in my previous answer without success.
