<s>[INST] 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.
 [/INST] 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
$.
</s>