Lower estimate for $(\frac{\ln(1+2x)}{\ln(1+x)}-1)(1+2x)^{1/2}$ where $x>0$

Question

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!

Answer 1

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!