I am trying to prove the following inequality:


$$\frac{y}{x}-1-\log\left(\frac{y}{x}\right)\geq \frac{1}{2}\frac{(x-y)^2}{x} \quad \forall x,y \in (0,1]$$


This inequality appears in the paper "[Scale-Free Adversarial Multi Armed Bandits](https://proceedings.mlr.press/v167/putta22a/putta22a.pdf)" as corollary 7, p. 8. The proof in the paper involves some geometric intuition, but the statement looks simple enough that it may have an algebraic proof. However, I have not been able to figure it out.



Proof of the inequality $\frac{y}{x}-1-\log\left(\frac{y}{x}\right)\geq \frac{1}{2}\frac{(x-y)^2}{x}$ when $x,y \in (0,1]$
$\newcommand\p\partial$Let $d(x,y)$ denote the difference between the left- and right-hand sides of the inequality in question. We want to show that $d(x,y)\ge0$ for $x$ and $y$ in $(0,1]$.


For such $x$ and $y$, we have $xy\,\p\_y d(x,y)=(x-y)(y-1)$ and $d(x,0+)=\infty$. So, without loss of generality (wlog), either $y=x$ (and then the inequality $d(x,y)\ge0$ is trivial) or $y=1$. Next, $2x^2\,\p\_x d(x,1)=-(1-x)^2\le0$ for $x\in(0,1]$. So, wlog $x=1=y$, and then the inequality $d(x,y)\ge0$ is trivial. $\quad\Box$


