= Solution
Normalize the <probability density function> using $u=t^c$:
$$
1=A_c\int_0^\infty t^{c-1}e^{-t^c}\,dt=\frac{A_c}{c}\int_0^\infty e^{-u}\,du,
\qquad \boxed{A_c=c.}
$$
The corresponding <Weibull distribution> has <survival function>
$$
S_c(v)=P(X_c\geq v)=\int_v^\infty ct^{c-1}e^{-t^c}\,dt=e^{-v^c},\qquad v\geq0.
$$
Since the distribution is continuous, the choices $>$ and $\geq$ have the same <probabilities>. The <conditional probability> is consequently
$$
P(X_c\geq s+t\mid X_c\geq t)=\frac{S_c(s+t)}{S_c(t)}=e^{-[(s+t)^c-t^c]}.
$$
For $s,t>0$ and $c>1$, the supplied inequality, applied to $s/t$, gives $(s+t)^c>t^c+s^c$. Taking exponentials yields
$$
\boxed{P(X_c\geq s+t\mid X_c\geq t)<e^{-s^c}=P(X_c\geq s).}
$$
This expresses the increasing <hazard function> $h_c(t)=ct^{c-1}$: surviving to a later age shortens the remaining lifetime in this <conditional distribution>. At $c=1$ equality holds, as expected from the <memoryless property> of the <exponential distribution>.
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