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Let $A,B,d\ge 1$ and suppose that $x\ge0$ satisfies $$ x^{\frac{d+1}{d}} \le Ax+B. \qquad(*) $$ I can show that $(*)$ implies the bound $$ x< d(A^d+B). \qquad(**) $$

Questions: (1) Can a better bound than $(**)$ be obtained? (2) Can such bounds be easily derived from some general "master" theorem?

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    $\begingroup$ Is the initial condition greater than or equal to $1$ only imposed on $d$, or on $A$ and $B$ too? $\endgroup$
    – Pietro Majer
    Commented Aug 23, 2019 at 6:40
  • $\begingroup$ on all 3 -- otherwise, the claim is false $\endgroup$
    – Aryeh Kontorovich
    Commented Aug 23, 2019 at 6:44
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    $\begingroup$ To improve a bit your bound you may multiply $(*)$ by $\lambda ^{\frac{ d+1}{d}}$, deducing a family of $(*)_\lambda$ for $\lambda x$, with $\lambda$ in the range of validity of your initial bound (that is, with corresponding $A_\lambda\ge1, B_\lambda\ge1$). Then you may optimize w.r.to $\lambda$ the corresponding bound on $x$ given by $(**)_\lambda$ $\endgroup$
    – Pietro Majer
    Commented Aug 23, 2019 at 6:57

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I assume $d$ is an integer. Let $x=y^d$, then the inequality takes form: $$y^{d+1} \leq Ay^d+B.$$ It holds for $y\leq Y$, where $Y$ is the largest zero of the polynomial $y^{d+1} - Ay^d-B$.

There is a number of known bounds for polynomial zeros. For example, Cauchy's bound gives $Y\leq 1+\max(A,B)$ implying that $x\leq (1+\max(A,B))^d$. Under scaling we can also get $$x\leq ((B/A)^{1/d}+A)^d.$$ Alternatively, Hölder's inequality for $k=d$ gives $$x\leq 1+(A^{d/(d-1)}+B^{d/(d-1)})^{d-1},$$ and so on.

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  • $\begingroup$ Is this bound better than (**)? $\endgroup$
    – Aryeh Kontorovich
    Commented Aug 23, 2019 at 8:28
  • $\begingroup$ If $B<<A$, then Yes. $\endgroup$
    – Max Alekseyev
    Commented Aug 23, 2019 at 8:39
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    $\begingroup$ Following this idea, and putting $y:=1/u$, the problem translates into a bound for the smallest root of a trinomial equation $u+u^q=c$, for which however there is a power series solution mathoverflow.net/questions/249060/… $\endgroup$
    – Pietro Majer
    Commented Aug 23, 2019 at 9:46
  • $\begingroup$ @PietroMajer: Yes, this is in the same venue as mathoverflow.net/q/32261 $\endgroup$
    – Max Alekseyev
    Commented Aug 23, 2019 at 17:45

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