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If $K=0$ or $S\le K$, the proposed sharp power-call inequality is immediate. Otherwise put $x=S/K>1$ and consider the ratioIts logarithmic derivative is $(1+\epsilon )/x-1/(x-1)$, whose sign is that of $\epsilon x-(1+\epsilon )$. The ratio decreases and then increases, with minimum at $x_{\ast }=(1+\epsilon )/\epsilon$. That minimum is $(1+\epsilon {)}^{1+\epsilon }/{\epsilon }^{\epsilon }$. Rescaling proves the sharp power-call inequalityThe constant is sharp because equality holds for $S=(1+\epsilon )K/\epsilon$ when $K>0$. This also explains the hinted minimization: with $a$ equal to that constant times $K^{\epsilon }$, the minimum of $s^{1+\epsilon }-as$ is $-aK$. Its denominator is ${\epsilon }^{\epsilon }$, including when reading the original PDF.

Take expected values in the sharp power-call inequality. If $M(1+\epsilon )$ is finite, then for every $K\ge 0$,The bound is independent of the strike, soCombined with the previous part, this relates finite moments to polynomial decay of expected European call option payoffs, while retaining the distinction at the moment threshold.