Related to New bound for Am-Gm of 2 variables we have :
Let $x\geq 5$ be a real number then we have : $$\sqrt{x}+x^{\frac{x}{x+1}}\geq \frac{x^2+1}{x+1}\Bigg(\frac{x^{\frac{x}{x+1}}}{x^{\frac{x^2}{x^2+1}}}+\frac{\sqrt{x}}{x^{\frac{x}{x+1}}}\Bigg)\geq x+1$$
For the LHS we have the following inequality for $x\geq 5$ :
$$\sqrt{x}+x^{\frac{x}{x+1}}- \frac{x^2+1}{x+1}\Bigg(\frac{x^{\frac{x}{x+1}}}{x^{\frac{x^2}{x^2+1}}}+\frac{\sqrt{x}}{x^{\frac{x}{x+1}}}\Bigg)< 1$$
So we compute the limit at infinity to get :
$$\lim_{x\to \infty}\sqrt{x}+x^{\frac{x}{x+1}}- \frac{x^2+1}{x+1}\Bigg(\frac{x^{\frac{x}{x+1}}}{x^{\frac{x^2}{x^2+1}}}+\frac{\sqrt{x}}{x^{\frac{x}{x+1}}}\Bigg)= 1$$
I can prove it using generalized Puiseux series at infinity and the Hospital rule we have :
$$x^{\frac{x}{x+1}}=x - \log(x) + \frac{\log(x) (\log(x) + 2)}{2 x} + O\Big(\frac{1}{x^2}\Big)$$
And
$$x^{\frac{x^2}{x^2+1}}=x - \frac{\log(x)}{x} + O\Big(\frac{1}{x^2}\Big)$$
And :
$$\frac{x^2+1}{x+1}=x - 1 + \frac{2}{x} - \frac{2}{x^2} + \frac{2}{x^3} + O\Big(\frac{1}{x^4}\Big)$$
But $$f(x)=\sqrt{x}+x^{\frac{x}{x+1}}$$ and $$g(x)= \frac{x^2+1}{x+1}\Bigg(\frac{x^{\frac{x}{x+1}}}{x^{\frac{x^2}{x^2+1}}}+\frac{\sqrt{x}}{x^{\frac{x}{x+1}}}\Bigg)$$
Are both strictly increasing $\forall x>5$ so we have prove it for a $q<x$ . Remains to prove that $q=5$.
Update :
Like this my reasoning is incomplete we need to prove that the functions $g(x)$ and $f(x)$ have no inflection point
End update
For the RHS .
It sufficient to prove for $x$ a large real number :
$$\sqrt{x}+x^{\frac{x}{x+1}}-1\geq x+1$$
I can prove it using derivatives and with the function :
$h(x)=\sqrt{x}+x^{\frac{x}{x+1}}-1-x+1$
So the real problem is when $x$ is relatively small .
My question :
How to solve it when $x$ is relatively small ?
Any helps is greatly appreciated.
Thanks a lot for all your contributions.
