= Solution
The sets of admissible <functions> are nested:
$$
\mathcal A_{\mathrm{clamped}}\subseteq\mathcal A_{\mathrm{pinned}}\subseteq\mathcal A_{\mathrm{free}}.
$$
All impose the same left conditions; the right height adds a restriction, and the right slope adds another. Minimizing the same <functional> over a smaller set cannot lower its minimum. Therefore
$$
\boxed{E_{\mathrm{free}}\leq E_{\mathrm{pinned}}\leq E_{\mathrm{clamped}}.}
$$
\b[The doubly clamped beam has the greatest minimum energy, and the free-ended beam the smallest.] No energy integrals are needed. In fact the inequalities are strict: the positive quadratic bending term makes the minimizer unique on the left-clamped space, and the three solutions in part (b) are different. The relevant principle is <monotonicity of a constrained minimum under added constraints>.
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