Lower estimate for $(\frac{\ln(1+2x)}{\ln(1+x)}-1)(1+2x)^{1/2}$ where $x>0$ I want to prove that:
$$\left(\frac{\ln(1+2x)}{\ln(1+x)}-1\right)(1+2x)^\frac{1}{2}\geq 1$$
where $x>0$. Any help appreciated. Thanks!
 A: Here's a not very beautiful solution.
As $x > 0$, it is equivalent to show that
$$\log\left(1 + \frac{x}{1 + x}\right)\sqrt{1 + 2 x} - \log(1 + x) \geq 0.$$
As LHS $\rightarrow 0$ as $x\rightarrow 0$, it is enough to show that the LHS is increasing as a function of $x$. That is, that its derivative is greater than $0$ for $x > 0$. Differentiating, we obtain the inequality
$$\frac{1 - \sqrt{1 + 2 x} + (1 + x) \log\left(1 + \frac{x}{1 + x}\right)}{(1 + x) \sqrt{1 + 2 x}} \geq 0.$$
It is of course enough to show that 
$$1 - \sqrt{1 + 2 x} + (1 + x) \log\left(1 + \frac{x}{1 + x}\right)\geq 0.$$
By the same argument (i.e., limit when $x\rightarrow 0$ is $0$, so it is enough to show LHS is increasing as a function of $x$),
differentiating, it is enough to show that
$$\frac{1}{1 + 2 x} - \frac{1}{\sqrt{1 + 2 x}} + \log\left(1 + \frac{x}{1 + x}\right) \geq 0.$$
By the same argument yet again (sorry!), it is enough to show that
$$\frac{1 - \sqrt{1 + 2 x} + x (3 + 2 x)}{(1 + x) (1 + 2 x)^{\frac{5}{2}}} \geq 0.$$
So it is enough to show that
$$1 - \sqrt{1 + 2 x} + x (3 + 2 x) \geq 0.$$
But this one is easy!
