Problem 11360
(American Mathematical Monthly, Vol.115, April 2008) Proposed by C. Lupu and T. Lupu (Romania).
Letf and g be continuous real-valued functions on [0, 1] satisfying the conditionR1
0 f g = 0. Show that
Z 1
0
f2 Z 1
0
g2≥ 4
Z 1 0
f Z 1
0
g
2
and
Z 1
0
f2
Z 1
0
g
2 +
Z 1
0
g2
Z 1
0
f
2
≥ 4
Z 1
0
f Z 1
0
g
2 .
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
Let
Z 1
0
f2= A, Z 1
0
g2= B, Z 1
0
f = a, Z 1
0
g = b, we will prove that
AB ≥ Ab2+ Ba2≥ 4a2b2.
By Cauchy-Schwarz inequality, B ≥ b2 and equality holds iff g is constant. In that case,R1
0 f g = 0 implies that a = 0 and the inequalities hold trivially. So we can assume that B > b2.
By Cauchy-Schwarz inequality, for any real t we have that Z 1
0
1 · Z 1
0
(f + tg)2≥
Z 1
0
[1 · (f + tg)]
2
that is, sinceR1
0 f g = 0,
A + Bt2≥ a2+ 2abt + b2t2 or
A ≥ sup
t∈R
{a2+ 2abt − (B − b2)t2}.
Since B > b2, then the second degree polynomial assume its maximum value for t = ab/(B − b2) and we have that
A ≥ a2+ 2ab ab
(B − b2)− (B − b2) a2b2
(B − b2)2 = a2+ a2b2 (B − b2) that is
AB ≥ Ab2+ Ba2. Finally, again by Cauchy-Schwarz inequality,
AB ≥ Ab2+ Ba2= Z 1
0
(bf + ag)2≥
Z 1
0
(bf + ag)
2
= (2ab)2= 4a2b2.
