= Solution
Write $F_X(x)=\mathbb P(X\leq x)$ and $F_Y(x)=\mathbb P(Y\leq x)$. For $0\leq x\leq\mu$, the factor $t/\mu$ is at most one throughout $0\leq t\leq x$, so
$$
F_Y(x)=\int_0^x\frac t\mu f(t)\,dt\leq\int_0^x f(t)\,dt=F_X(x).
$$
For $x>\mu$, compare the tails instead: $t/\mu\geq1$ throughout $t\geq x$, giving
$$
1-F_Y(x)=\int_x^\infty\frac t\mu f(t)\,dt\geq\int_x^\infty f(t)\,dt=1-F_X(x).
$$
Rearranging proves
$$
\boxed{\mathbb P(Y\leq x)\leq\mathbb P(X\leq x)\quad(x\geq0).}
$$
This is <stochastic domination> of $X$ by its <size-biased distribution>. The proof needs only the finite positive <mean>, not the finite <variance> assumed in (ii).
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