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Prove it.","posed_since":"1976","precise":"For $-\\mathrm{d}^2/\\mathrm{d}x^2 + V$ on $L^2(R)$, with $L_{\\gamma,1}$ the sharp constant in $\\sum_j \\left|E_j\\right|^\\gamma \\le L_{\\gamma,1} \\int V_-^{(\\gamma + 1/2)} dx$, prove $L_{\\gamma,1} = L^1_{\\gamma,1}$ for $1/2 < \\gamma < 3/2$, where $L^1_{\\gamma,1}$ is the sharp constant over potentials with exactly one negative eigenvalue. Known endpoints: equality at $\\gamma = 1/2$ (Hundertmark, Lieb, Thomas 1998) and $L_{\\gamma,1} = L^{\\mathrm{cl}}_{\\gamma,1}$ for $\\gamma \\ge 3/2$.","problem_ref":"4488e6f955ebf7c16bc621b98839387488a20933b532c363830bee662d9a397b","references":"","settled_by":"A proof that multi-bound-state potentials never beat the single-bound-state value for $1/2 < \\gamma < 3/2$, or a numerical-plus-rigorous counterexample.","status_note":"The 2021 Frank-Gontier-Lewin disproof applies in $d = 1$ only for $\\gamma > 3/2$, so the conjecture remains consistent with current results; at $\\gamma = 1$ the best bound is $L_{1,1}/L^{\\mathrm{cl}}_{1,1} \\le 1.456$ against the conjectured $2/\\sqrt{3}$, about 1.155 (Frank, Hundertmark, Jex, Nam 2021).","title":"One-bound-state optimality in one dimension for $1/2 < \\gamma < 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