= Solution
As in part (c), every admissible <transport map> has nonnegative displacement $D(x)=T(x)-x$ with $\int_0^1D(x)\,dx=1$. The square root is a <strictly concave function>, so the reversed <Jensen inequality> gives
$$
\mathbb M(T)=\int_0^1\sqrt{D(x)}\,dx\leq\sqrt{\int_0^1D(x)\,dx}=1.
$$
The map $T^\dagger(x)=x+1$ is admissible and attains one. Therefore
$$
\boxed{\max_{T_\#\mu=\nu}\mathbb M(T)=1,\qquad T^\dagger(x)=x+1.}
$$
Thus “worst” means largest cost among admissible <transport maps>. Equality in the <Jensen inequality> for a <strictly concave function> forces $D(x)$ to be constant almost everywhere, so this maximizer is unique up to a $\mu$-null set. For comparison, the admissible reflection $T(x)=2-x$ has smaller cost
$$
\int_0^1\sqrt{2-2x}\,dx=\frac{2\sqrt2}{3}<1.
$$
The upper bound also holds for every <transport plan>.
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