= 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 ratio
$$
f(x)=\frac{x^{1+\varepsilon}}{x-1}.
$$
Its logarithmic derivative is $(1+\varepsilon)/x-1/(x-1)$, whose sign is that of $\varepsilon x-(1+\varepsilon)$. The ratio decreases and then increases, with minimum at $x_*=(1+\varepsilon)/\varepsilon$. That minimum is $(1+\varepsilon)^{1+\varepsilon}/\varepsilon^\varepsilon$. Rescaling proves the <sharp power-call inequality>
$$
\boxed{S^{1+\varepsilon}\ge\frac{(1+\varepsilon)^{1+\varepsilon}}{\varepsilon^\varepsilon}K^\varepsilon(S-K)_+.}
$$
The constant is sharp because equality holds for $S=(1+\varepsilon)K/\varepsilon$ when $K>0$. This also explains the hinted minimization: with $a$ equal to that constant times $K^\varepsilon$, the minimum of $s^{1+\varepsilon}-as$ is $-aK$. Its denominator is $\varepsilon^\varepsilon$, including when reading the original PDF.
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