Does $1 + \frac{1}{x} + \sqrt{\frac{2x}{x + 1}},$ have a global minimum? Does the following function have a global minimum:
$$1 + \frac{1}{x} + \sqrt{\frac{2x}{x + 1}},$$
where $x \in \mathbb{N}$?
I tried using WolframAlpha, but it appears to give an inconsistent result.
 A: $$1 + \frac{1}{x} + \sqrt{\frac{2x}{x + 1}}$$
$$=1 + \frac{1}{x} + \sqrt{\frac{2}{1 + \frac{1}{x}}}$$
$$=1 + \frac{1}{x} + \frac{1}{2}\sqrt{\frac{2}{1 + \frac{1}{x}}}+\frac{1}{2}\sqrt{\frac{2}{1 + \frac{1}{x}}}$$
Applying A.M. G.M. we have,
$$1 + \frac{1}{x} + \frac{1}{2}\sqrt{\frac{2}{1 + \frac{1}{x}}}+\frac{1}{2}\sqrt{\frac{2}{1 + \frac{1}{x}}}\geq3.((1 + \frac{1}{x})(\frac{1}{2}\sqrt{\frac{2}{1 + \frac{1}{x}}})^2)^{\frac{1}{3}}$$
$$3.((1 + \frac{1}{x})(\frac{1}{2}\sqrt{\frac{2}{1 + \frac{1}{x}}})^2)^{\frac{1}{3}}=3.(\frac{1}{2})^{1/3}$$
Equality holds when $\displaystyle 1 + \frac{1}{x}=\frac{1}{2}\sqrt{\frac{2}{1 + \frac{1}{x}}}$
Squaring both sides we get,
$$(1 + \frac{1}{x})^{3}=\frac{1}{2}$$
$$\Rightarrow 1 + \frac{1}{x}=\frac{1}{2^{1/3}}$$
$$\Rightarrow \frac{1}{2^{1/3}}-1=\frac{1}{x}$$
$$\Rightarrow x=\frac{2^{1/3}}{1-2^{1/3}}$$
Now check the two nearest integers to x and compare the values of the expression at those values and the min. will global minimum . 
A: The derivative of the expression is
$$
\frac{1}{x(x+1)}\sqrt{\frac{x}{2(x+1)}}-\frac{1}{x^2},
$$
which is strictly negative for $x>0$. Hence, the function is always decreasing.
When $x<-1$, you can show that the derivative is zero at $x=-2-\sqrt[3]{2}-2^{2/3}\approx -4.84732$. See WolframAlpha here and here.
Thus, a local minima occurs at $x=-4$ or $x=-5$. There is no local minima for $x>0$.
A: Since the function
$$f(x) = 1 + \frac{1}{x} + \sqrt{\frac{2x}{x + 1}}$$
is always decreasing (as pointed out by Daryl),
we have
$$f(x) > \lim_{x \rightarrow \infty}{f(x)} = 1 + \sqrt{2}.$$
