Problem 11933
(American Mathematical Monthly, Vol.123, October 2016)
Proposed by J. M. Pacheco and A. Plaza (Spain).
For positive integern, let Hn =Pn
k=11/k. Prove Z 1
0
1 x + 1dx ·
Z 1 0
x + 1
x2+ x + 1dx · · · Z 1
0
xn−2+ · · · + x + 1
xn−1+ · · · + x + 1dx ≥ 1 Hn.
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
Solution. We first note that Z 1
0
(1 + x + · · · + xn−1) dx =
nX−1 k=0
1
k + 1 = Hn.
We show the desired inequality by induction on n.
Base case. For n = 1 the product on the left-hand side is empty and therefore it is equal to 1 = 1/H1. Induction step. Let n ≥ 2 then, by the induction hypothesis,
Z 1 0
1 x + 1dx·
Z 1 0
x + 1
x2+ x + 1dx · · · Z 1
0
xn−2+ · · · x + 1
xn−1+ · · · + x + 1dx ≥ 1 Hn−1
Z 1 0
xn−2+ · · · + x + 1 xn−1+ · · · + x + 1dx
?
≥ 1 Hn. Hence, it suffices to show that
Hn Z 1
0
xn−2+ · · · + x + 1
xn−1+ · · · + x + 1dx ≥ Hn−1, that is,
Z 1
0
(1 + x + · · · + xn−1) dx · Z 1
0
xn−2+ · · · + x + 1 xn−1+ · · · + x + 1dx ≥
Z 1
0
(1 + x + · · · + xn−2) dx
which holds by Chebyshev Integral Inequality because f (x) := 1 + x + · · · + xn−1 is increasing in [0, 1],
g(x) := xn−2+ · · · + x + 1
xn−1+ · · · + x + 1 = 1 − 1 f (1/x) is decreasing in [0, 1], and f (x) · g(x) = 1 + x + · · · + xn−2.
