# Show $\log\left(\frac{x-y}{x}\right)-2\sqrt{\frac{x-y}{x}}<\log\left(\frac{x+y}{x}\right)-2\sqrt{\frac{x+y}{x}}$ if $x\geq 5$ and $1\leq y\leq x-2$

Assume that all logarithms are natural.

Let $$x$$ and $$y$$ be integers that satisfy $$x \geq 5$$ and $$1 \leq y \leq x-2$$. I am trying to show that $$\log\left( \frac{x-y}{x}\right)-2\sqrt{\frac{x-y}{x}} < \log\left( \frac{x+y}{x}\right)-2\sqrt{\frac{x+y}{x}}.$$

This is equivalent to showing that $$\log\left( \frac{x-y}{x+y}\right)<2\left(\sqrt{\frac{x-y}{x}}-\sqrt{\frac{x+y}{x}}\right).$$

My first inclination was to apply the well known inequality, $$\log(z)\leq 2(\sqrt{z}-1)$$ if $$z>0$$, to the left side by letting $$z=\frac{x-y}{x+y}$$, but this did not help me.

I also tried double induction on $$(x,y)$$ (since $$x$$ and $$y$$ are integers) but I could not finish it.

I appreciate any help, thank you.

The inequality $$\log\left( \frac{x-y}{x+y}\right)<2\left(\sqrt{\frac{x-y}{x}}-\sqrt{\frac{x+y}{x}}\right)$$ holds for all real numbers $$x, y$$ satisfying $$0 < y < x$$.

With the substitution $$t=y/x$$ this is equivalent to $$\tag{*} \log\left( \frac{1+t}{1-t}\right)>2\left(\sqrt{1+t}-\sqrt{1-t}\right)$$ for $$0 < t < 1$$. This suggests to investigate the function $$f(t) = \log\left( \frac{1+t}{1-t}\right) -2\left(\sqrt{1+t}-\sqrt{1-t}\right) \, .$$ Then $$f(0) = 0$$, and for $$0 < t < 1$$ is $$f'(t) = \frac{2}{1-t^2} - \left( \frac{1}{\sqrt{1+t}} + \frac{1}{\sqrt{1-t}} \right) \\ = \frac{2 - \sqrt{1-t^2}\left(\sqrt{1+t} + \sqrt{1-t}\right)}{1-t^2}$$ and that is positive, because $$\sqrt{1-t^2} \cdot (\sqrt{1+t} + \sqrt{1-t}) < \sqrt{1+t} + \sqrt{1-t} \le 2$$ as can be verified easily.

So $$f$$ is strictly increasing on $$[0, 1)$$, and that proves the inequality $$(*)$$.

Try working with the inequalities on $$x$$ and $$y$$. We are given $$x\in[5,\infty)$$ and $$y\in[1,x-2]$$. First, we find $$x-y\in[2,x-1]$$ and $$x+y\in[6,2x-2]$$. For a fraction, we can find the minimum by maximizing the denominator and minimizing the numerator, for instance $$\frac{x-y}{x+y}\in[\frac{2}{2x-2},\frac{x-1}{6}]$$.

Then, $$\log(z)$$ and $$\sqrt(z)$$ are strictly increasing functions so we can plug in the minimum and maximum values for their arguments. So it will suffice to show $$\log\left(\frac{x-1}{6}\right)<2\left(\sqrt{\frac{x-1}{5}}-\sqrt{\frac{6}{x}}\right)$$

We can split up the inequality to find the inequality you mentioned:

$$\log\left(\frac{x-1}{5}\cdot\frac{5}{6}\right)<2\left(\sqrt{\frac{x-1}{5}}-1\right)+2\left(1-\sqrt{\frac{6}{x}}\right)$$ $$\log\left(\frac{x-1}{5}\right)+\log\left(\frac{5}{6}\right)<2\left(\sqrt{\frac{x-1}{5}}-1\right)+2\left(1-\sqrt{\frac{6}{x}}\right)$$

And extracting the $$\log(z)\le 2(\sqrt(z)-1)$$ identity we need to show:

$$\log\left(\frac{5}{6}\right)<2\left(1-\sqrt{\frac{6}{x}}\right)$$

Which is clearly true because the lhs is negative and the rhs is positive for the allowed values of $$x$$.

Here's a proof by comparing the power series.

Eliding some computation,

$$\log\left( \frac{1+t}{1-t}\right) =2\sum_{n=1}^{\infty} \dfrac{t^{2n-1}}{2n-1}$$.

$$\sqrt{1+t}-\sqrt{1-t} =2\sum_{n=1}^{\infty} \binom{1/2}{2n-1}t^{2n-1} =2\sum_{n=1}^{\infty} \dfrac{1}{2^{4n-3}(2n-1)}\binom{4n-4}{2n-2}t^{2n-1}$$

So we want $$\dfrac{1}{2^{4n-3}(2n-1)}\binom{4n-4}{2n-2} \lt \dfrac{1}{2n-1}$$

or $$\dfrac{1}{2^{4n-3}}\binom{4n-4}{2n-2} \lt 1$$

or, with $$2n-2=m$$, $$\dfrac{1}{2^{2m+1}}\binom{2m}{m} \lt 1$$.

But $$(1+1)^{2m} =2^{2m}$$ and, for $$m \ge 1$$, $$(1+1)^{2m} =\sum_{k=0}^{2m}\binom{2m}{k} \gt \binom{2m}{m}$$ so $$\dfrac{1}{2^{2m+1}}\binom{2m}{m} \lt \dfrac12$$.