${2 x^{1-y}(1 - x^y) \over y (1-x)(1+x^{1-y})} \leq 1$ for $0<x<1, 0<y<1$

Question

The title says it all. I'd like to prove the inequality $${2 x^{1-y}(1 - x^y) \over y (1-x)(1+x^{1-y})} \leq 1$$ for $0<x<1, 0<y<1$. I'd be happy just showing that the left hand side is bounded. Mathematica seems to believe that the inequality holds, and some formal asymptotics also suggest it. For example, fix $y$, and send $x$ to $1$. Writing $x^y = e^{y \log x} = 1 + y \log x + O((\log x)^2)$, we obtain that $${1-x^y \over 1-x} \leq {y \log x + O((\log x)^2) \over \log x} = y + O(\log x) \to y$$ as $x \to 1$. Thus, for fixed $y$, the left hand side converges to $1$ as $x \to 1$. Similarly, for fixed $x$, as $y \to 0$, $1 - x^y = -y \log(x) + O(y^2)$, and so as $y \to 0$ the expression on the left converges to $${2 x \log x \over (1-x)(1+x)}.$$ Since $\log x \leq 1-x$ for all $x \in (0,1)$ and $2x \leq 1 + x$ for all $x \in (0,1)$, this is bounded by $1$.

Answer 1

Rewrite our inequality in the following form.

$$\frac{2x+y-xy}{2-y+xy}\geq x^{1-y}$$ or $$\ln\frac{2x+y-xy}{2-y+xy}\geq \ln{x^{1-y}}$$ or $f(x)\geq0,$ where $$f(x)=\ln(2x+y-xy)-\ln(2-y+xy)-(1-y)\ln{x}.$$ But $$f'(x)=\frac{2-y}{2x+y-xy}-\frac{y}{2-y+xy}-\frac{1-y}{x}=$$ $$=-\frac{y(1-y)(2-y)(x-1)^2}{x(2-y+xy)(2x+y-xy)}<0,$$ which says that $f$ decreases.

Thus, since $x<1$, we obtain: $$f(x)>f(1)=0$$ and we are done!

Answer 2

The Hermite-Hadamard inequality states that for convex $f$, $$\int_0^1 f(a+t(b-a))d t\leq \frac{f(a)+f(b)}{2}\text{.}$$ $f(x)=yx^{y-1}$ is convex for $0<y<1$.