If \Phi:[0,\infty )\rightarrow \mathbb{R} Φ : [ 0 , ∞ ) → R is convex and continuous with \Phi (0)=0 Φ ( 0 ) = 0 and if q\in (1,\infty ) q ∈ ( 1 , ∞ ) , q^{\prime }:=\frac{q}{q-1} q ′ : = q q − 1 , we first prove that the inequality \Phi \left( \int_{0}^{\infty }f(r)dr\right) \leq C\int_{0}^{\infty }f(r)\Phi ^{\prime }(r^{1/q^{\prime }})dr Φ ( ∫ 0 ∞ f ( r ) d r ) ≤ C ∫ 0 ∞ f ( r ) Φ ′ ( r 1 / q ′ ) d r for every f\in L^{q}(0,\infty) f ∈ L q ( 0 , ∞ ) , f\geq 0 f ≥ 0 with ||f||_{q}\leq 1 ∣ ∣ f ∣ ∣ q ≤ 1 holds when C=1 C = 1 . In general, both sides may be \pm \infty ± ∞ . Related inequalities for f\in L^{1}(\mathbb{R}^{N})\cap L^{q}(\mathbb{R}^{N}) f ∈ L 1 ( R N ) ∩ L q ( R N ) , f\neq 0 f ≠ 0 are derived. This inequality is independent of Jensen's inequality and, when q=\infty q = ∞ , it is an elaboration on an inequality of Steffensen which was discussed elsewhere by the author. The next goal of the paper is to identify the range of the admissible constants C C and, in particular, to characterize the optimal constant when \Phi \geq 0 Φ ≥ 0 or \Phi \leq 0 Φ ≤ 0 . It turns out that C=1 C = 1 is “almost always” optimal, at least in a restricted sense, but not always when q<\infty q < ∞ : Given q q , the admissible constants lie on an interval containing 1 1 whose left (right) endpoint is the supremum (infimum) of a function defined on some (left/right dependent) subset of \mathbb{R}^{2} R 2 . If q=2 q = 2 , these extrema can be calculated in a number of examples. Among other things, this reveals that C=1 C = 1 need not be optimal when \Phi \geq 0 Φ ≥ 0 and \Phi _{+}^{\prime }(0)=0 Φ + ′ ( 0 ) = 0 or when \Phi \leq 0 Φ ≤ 0 and \Phi _{+}^{\prime} (0)=-\infty Φ + ′ ( 0 ) = − ∞ .