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/ ( & ( , !. 2 1 0 / 5 4 3 d ?. (a) f′′(x) ≤ 0 for x ≥ 0 (b) Since t2/2 is convex we have t2/2 ≥ x2/2x(t−x) = xt−x2/2 This is the general inequality g(t) ≥ g(x)g′(x)(t−x), which holds for any differentiable convex function, applied to g(t) = t2/2 Another (easier?) way to establish t2/2 ≤ −x2/2xt is to note that t2/2x2/2−xt = (1/2)(x−t)2 ≥ 0 Now just move x2/2−xt to the other side.