Proving an inequality involving logarithms in two dimensions I'm trying to show that the following inequality holds for $x, y \geq 0$,
$x\log(1 + \frac{y}{x}) + y \log(1 + \frac{x}{y}) \geq \frac{4\log(2) x y}{x + y}$.
Now, for exponentials the following is known to be true:
$\exp(\frac{xy}{x + y}) < (1 + \frac{x}{y})^y$ for $x, y > 0$,
which gives (applying logarithms and adding twice)
$x\log(1 + \frac{y}{x}) + y \log(1 + \frac{x}{y}) > \frac{2 x y}{x + y}$.
However, this doesn't give a tight enough inequality since $4\log(2) > 2$. Any help would be greatly appreciated.
 A: Dividing the both sides of 
$$x\log\left(1+\frac yx\right)+y\log\left(1+\frac xy\right)\ge \frac{4\log(2)xy}{x+y}$$
by $x\gt 0$ and setting $t=\frac yx$ give
$$\log (1+t)+t\log\left(1+\frac 1t\right)\ge \frac{4\log(2)}{\frac 1t+1}$$
Multiplying the both sides by $\frac 1t+1\gt 0$ gives
$$\left(\frac 1t+1\right)\log(1+t)+(1+t)\log\left(1+\frac 1t\right)\ge 4\log (2)$$
Let $f(t)$ be the LHS.
We want to prove that $f(t)\ge 4\log(2)$ for $t\gt 0$.
Now,
$$f'(t)=\log\left(\frac 1t+1\right)-\frac{\log(1+t)}{t^2},\quad f''(t)=\frac{2\log(t+1)-t}{t^3}$$
Let $g(t)=2\log(t+1)-t$. 
Then,
$$g'(t)=\frac{2}{t+1}-1=\frac{1-t}{t+1}$$
So, $g(t)$ is increasing for $0\lt t\lt 1$ and is decreasing for $t\gt 1$ with $\lim_{t\to 0^+}g(t)=0$ and $\lim_{t\to\infty}g(t)=-\infty$.
Letting $t=\alpha\gt 1$ be the only real solution of $g(t)=0$, we have that $f'(t)$ is increasing for $0\lt t\lt\alpha$ and is decreasing for $t\gt\alpha$.
Since
$$\lim_{t\to 0^+}f'(t)=-\infty,\quad \lim_{t\to\infty}f'(t)=0,\quad f'(1)=0$$
we see that $f(t)$ is decreasing for $0\lt t\lt 1$ and is increasing for $t\gt 1$.
Thus, we have that
$$f(t)\ge f(1)=4\log(2)$$
