Prove Positivity of a Function Involving $\log$. I want to prove
The function $f:(0,\infty)^2 \to \mathbb{R}$ defined by
\begin{equation}\label{key}
   f(x,y) =  \ln \left( \dfrac{1+y}{1+ x}\right) + \dfrac{x}{1+ x} \ln \left( \dfrac{x}{y}\right)
  \end{equation}
is non-negative.
My attempt was to compute   $\dfrac{\partial f}{\partial y}$ and then see where the zeros of the function are.
I think that works fine, but I was wondering if there was a slicker way to do it.
 A: Writing the logarithms as integrals over $1/t$ gives
$$
 (x+1)f(x, y) = (x+1) \int_{1+x}^{1+y} \frac{dt}{t} + x \int_y^x \frac{dt}{t} = \int_x^y \left(\frac{x+1}{t+1} - \frac xt\right) \, dt \\
= \int_x^y \frac{t-x}{t(t+1)} \, dt  = \int_0^{y-x} \frac{s}{(s+x)(s+x+1)} \, ds\, ,
$$
and that is non-negative: The integrand is $\ge 0$ if $x \le y$, and $\le 0$ otherwise.
One can also see that equality holds exactly if $x=y$.
A: Here is an elementary proof without calculus: Combine the logarithms and obtain
$$
f(x,y)=\ln\left[\frac{1+y}{1+x}\cdot\left(\frac{x}{y}\right)^{\frac{x}{1+x}}\right].
$$
We have to show that the argument of the logarithm is always $\geq 1$. This is equivalent to
$$
\frac{1+y}{1+x}\geq \left(\frac{y}{x}\right)^{\frac{x}{1+x}}.
$$
However, using Bernoulli's inequality and keeping in mind that $0<\frac{x}{1+x}\leq 1$ and $\frac{y-x}{x}>-1$, we have
$$
\left(\frac{y}{x}\right)^{\frac{x}{1+x}}=\left(1+\frac{y-x}{x}\right)^{\frac{x}{1+x}}\leq 1+\frac{y-x}{x}\cdot\frac{x}{1+x}=1+\frac{y-x}{1+x}=\frac{1+y}{1+x}
$$
and so we are done!
