When $\left(2+\frac{x}{y}+\frac{y}{x}\right)^x$ is larger that $(1+x)^{1+x}$?

Question

I need to find when:

$$\left(2+\frac{x}{y}+\frac{y}{x}\right)^x>(1+x)^{1+x}$$

given that $0<y<x<1$. I cannot manage to find a necessary and sufficient condition.

Answer 1

$$2+\frac{x}{y}+\frac{y}{x}\ge2+2=4$$ with equality iff $x=y$.

$(1+x)^{1+\frac{1}{x}}$ is an increasing function for $x\ge-1$, since its derivative is $$(1+x)^{1+1/x}(1-\ln(1+x)/x)/x\ge0.$$ (Note $\ln(1+x)\le x$.) So on the given range it takes its maximum value at $x=1$ of $2^2$. Hence the two expressions meet at the value of 4 when $x=1=y$.

Answer 2

Take logs and you're trying to prove $$x \ln \left( 2 + \frac{x}{y} + \frac{y}{x} \right) > (1+x) \ln (1+x)$$

Using Barry Cipra's hint that $\frac{x}{y} + \frac{y}{x} \geq 2$ (with equality iff $x = y$), it suffices to show that $$x \ln 4 > (1+x) \ln (1+x)$$ and as $(1+x) \ln (1+x)$ is a convex function, it suffices to show that equality holds at $x = 0$ and $x = 1$ to show that the inequality holds for all $x \in (0, 1)$ and all positive values of $y$.