Does $1 + \frac{1}{x} + \sqrt{\frac{2x}{x + 1}},$ have a global minimum?

$$1 + \frac{1}{x} + \sqrt{\frac{2x}{x + 1}}$$

$$=1 + \frac{1}{x} + \sqrt{\frac{2}{1 + \frac{1}{x}}}$$

$$=1 + \frac{1}{x} + \frac{1}{2}\sqrt{\frac{2}{1 + \frac{1}{x}}}+\frac{1}{2}\sqrt{\frac{2}{1 + \frac{1}{x}}}$$

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 .

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 $x>0$.