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.

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 $x$.
