# Introduction

We look at inequalities of the form $$\label{eq:first} x ~\le~ \alpha + \beta x^\kappa$$ for $$\alpha \ge 0$$, $$\beta \ge 0$$ and $$\kappa \in (0,1)$$. Intuitively, this inequality states that $$x$$ cannot be too large, since $$x$$ has to be below a smaller power of itself. But how large can $$x$$ still be? Well, obviously, $x ~\le~ v \qquad\text{where}\qquad v ~=~ \sup \left\{ x \middle| x \le \alpha + \beta x^\kappa \right\} .$ Now $$v$$ satisfies $$\eqref{eq:first}$$ with equality, i.e. $$v = \alpha + \beta v^\kappa$$. But, unfortunately, that equality does not generally have an analytic solution. So we are interested in tight and tractable upper bounds on $$v$$.

To set the stage, let’s start with a simple lower bound on $$v$$. If we set $$\alpha = 0$$, we can indeed solve for equality in $$\eqref{eq:first}$$, yielding $v ~\ge~ \beta^{\frac{1}{1-\kappa}} .$ We are hence (by monotonicity of $$v$$ in $$\alpha$$) looking for an expression that is a little larger than this.

# A Solution

Since $$\kappa \in [0,1]$$, the function $$x \mapsto x^\kappa$$ is concave. The main idea will be to linearise it around some point $$y \ge 0$$, i.e. $x^\kappa ~\le~ y^\kappa + (x-y) \kappa y^{\kappa-1} ~=~ (1-\kappa) y^{\kappa} + \kappa x y^{\kappa-1}$ and hence $v ~=~ \alpha + \beta v^\kappa ~\le~ \alpha + \beta \left( (1-\kappa) y^{\kappa} + \kappa v y^{\kappa-1} \right) .$ This inequality is trivial for $$y \le (\beta \kappa)^{\frac{1}{1-\kappa}}$$. For larger $$y$$ we may reorganise it to $v ~\le~ \frac{ \alpha + \beta (1-\kappa) y^{\kappa} }{ 1 - \beta \kappa y^{\kappa-1} } .$ It then remains to pick $$y$$. Optimising the above expression in $$y$$ will not work, as it would solve $$\eqref{eq:first}$$ exactly. But let us plug in the following reasonable choice (which would be optimal were $$\alpha=0$$) $y ~=~ \arg\min_{y \ge 0} ~ \frac{ \beta (1-\kappa) y^{\kappa} }{ 1 - \beta \kappa y^{\kappa-1} } ~=~ \beta^{\frac{1}{1-\kappa}}$ and the final bound becomes $v ~\le~ \frac{ \alpha }{ 1 - \kappa } + \beta^{\frac{1}{1-\kappa}} .$ That looks useful, especially for small $$\alpha$$ (which happens to be the case that sparked my interest). Moreover, this bound is tight in the following way. If we write $$v(\alpha)$$ for $$v$$ regarded as a function of $$\alpha$$, then $$v(0) = \beta^{\frac{1}{1-\kappa}}$$ and $$v'(0) = \frac{1}{1-\kappa}$$. So the expression above is the Taylor expansion of $$v(\alpha)$$ around $$\alpha = 0$$.