(1)Problem 11780 (American Mathematical Monthly, Vol.121, May 2014) Proposed by C

Problem 11780

(American Mathematical Monthly, Vol.121, May 2014) Proposed by C. Lupu (USA) and T. Lupu (Romania).

Let f be a positive-valued, concave function on [0, 1] Prove that 3

4

Z 1 0

f (x) dx

²

≤ 1 8+

Z 1 0

f³(x) dx.

Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.

We note that, for t ≥ 0,

t³−3 4t²+ 1

16 = (4t + 1)(2t − 1)²

16 ≥ 0.

Hence, since f is non-negative, it follows that Z 1

0



f³(x) −3

4f²(x) + 1 16



dx ≥ 0,

that is, by Cauchy-Schwarz inequality, Z 1

0

f³(x) dx + 1 16 ≥3

4 Z 1

0

f²(x) dx ≥3 4

Z 1 0

f (x) dx

²

which is stronger than the required inequality. The concavity property is not necessary.