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.
 A: 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 $(*)$.
A: 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 $x$.
A: 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
$.
