Problem 12023
(American Mathematical Monthly, Vol.125, February 2018) Proposed by V. Mikayelyan (Armenia).
Letα be a positive real number. Prove Z π
0
xα−2sin(x) dx ≥ πα(α + 6) α(α + 2)(α + 3).
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
Solution. Since the function f (x) := (1 − cos(x))/x2 is concave in (0, π/2] (see below), it follows that the tangent line at π/2 stays above the graph of f : for x ∈ (0, π/2],
1 − cos(x)
x2 = f (x) ≤ f′(π/2)(x − π/2) + f (π/2) = 2(6 − π)
π2 −4(4 − π)x π3 . Hence, for s ∈ [0, 1/2],
cos(πs) ≥ 1 − 2(6 − π)s2+ 4(4 − π)s3≥1 − 6s2+ 4s3, and
sin(πt) = π Z t
0
cos(πs)ds ≥ π Z t
0
(1 − 6s2+ 4s3)ds = π(t − 2t3+ t4) := p(t).
Moreover, the symmetries sin(πt) = sin(π(1 − t)) and p(t) = p(1 − t) imply that the above inequality holds in [0, 1]. Then
Z π
0
xα−2sin(x) dx = πα−1 Z 1
0
tα−2sin(πt) dt
≥πα−1 Z 1
0
tα−2p(t) dt
= πα Z 1
0
(tα−1−2tα+1+ tα+2) dt
= πα 1 α− 2
α + 2+ 1 α + 3
= πα(α + 6) α(α + 2)(α + 3).
Proof of the concavity of f (x) = (1 − cos(x))/x2 in (0, π/2]. Let g(x) := x4f′′(x) then
g(x) = (x2−6) cos(x) − 4x sin(x) + 6, g′(x) = −(x2−2) sin(x) − 2x cos(x), g′′(x) = −x2cos(x).
Now for x ∈ [0, π/2], g′′(x) ≤ 0 and, together with g′′(0) = 0, it follows that g′(x) ≤ 0. Similarly, g(x) ≤ 0 and we may conclude that f si concave in (0, π/2].
