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.


3 Answers 3


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:


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 $.


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