Display (37): $D/q>h/\rho$ implies $E[\min(U,D/q)]/E[U]\ge E[\min(U,h/\rho)]/E[U]$ and $E[\min(U,D/q)]/E[U]>h/\rho$ (corrected: second inequality dropped)

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Display (37): $D/q>h/\rho$ implies $E[\min(U,D/q)]/E[U]\ge E[\min(U,h/\rho)]/E[U]$ and $E[\min(U,D/q)]/E[U]>h/\rho$ (corrected: second inequality dropped) · Prove2Me