Problem 12229
(American Mathematical Monthly, Vol.128, January 2021) Proposed by M. Omarjee (France).
Let f: [0, 1] → R be a function that has a continuous second derivative and that satisfies f (0) = f (1) andR1
0 f(x) dx = 0. Prove
30240
Z 1
0
xf(x) dx
2
≤ Z 1
0
(f′′(x))2dx.
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
Solution. By integrating by parts twice, we have that 12
Z 1
0
xf(x) dx = 12 Z 1
0
xf(x) dx − 6 Z 1
0
f(x) dx = Z 1
0
(12x − 6)f (x) dx = Z 1
0
f(x) d(6x2−6x + 1)
=(6x2−6x + 1)f (x)1
0− Z 1
0
(6x2−6x + 1)f′(x) dx
= (f (1) − f (0)) − Z 1
0
(6x2−6x + 1)f′(x) dx
= − Z 1
0
f′(x) d((2x3−3x2+ x))
= −(2x3−3x2+ x)f′(x)1
0+ Z 1
0
(2x3−3x2+ x)f′′(x) dx
= Z 1
0
(2x3−3x2+ x)f′′(x) dx.
Hence, by Cauchy-Schwarz inequality,
12
Z 1
0
xf(x) dx
2
≤ Z 1
0
(2x3−3x2+ x)2dx · Z 1
0
(f′′(x))2 dx.
Finally, sinceR1
0(2x3−3x2+ x)2dx=2101 , and 122·210 = 30240, it follows that 30240
Z 1
0
xf(x) dx
2
≤ Z 1
0
(f′′(x))2dx.
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