= Solution
{c}
Let $d=y-x$ and take $0<t\leq1$. Rearrange the defining inequality for a <strongly convex function>, applied to $x+td$:
$$
f(y)\geq f(x)+\frac{f(x+td)-f(x)}{t}+\frac\gamma2(1-t)\|d\|^2.
$$
Because $f$ is <differentiable> at $x$, its <directional derivative> in direction $d$ is
$$
\lim_{t\downarrow0}\frac{f(x+td)-f(x)}{t}=\nabla f(x)^Td.
$$
Taking the limit in the rearranged inequality gives
$$
\boxed{f(y)\geq f(x)+\nabla f(x)^T(y-x)+\frac\gamma2\|y-x\|^2.}
$$
This is the supporting <affine function> inequality strengthened by a positive squared <Euclidean distance>. \b[Differentiability at $x$ alone suffices]; no <derivative> at $y$ is used.
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