Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 39 / 2 / c / Solution

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.