American Mathematical Monthly Problem 11417 by Cezar Lupu and Tudorel Lupu
Letf : [0,1]→Rbe a continuously differentiable function such that
23
R
13
f(x)dx= 0. Prove the inequality
1
Z
0
(f0(x))2dx ≥27
1
Z
0
f(x)dx
2
.
Solution (by Darij Grinberg).
First, we somewhat simplify the problem by a linear substitution: We define a function g : [0,3] → R by g(x) = fx
3
for all x ∈ [0,3]. Then, g is a continuously differentiable function satisfying
2
R
1
g(x)dx= 0, and we have to prove that
3
Z
0
(g0(x))2dx≥
3
Z
0
g(x)dx
2
. (11417.1)
In order to prove this, we first show that
1
Z
0
(g0(x))2dx≥3
1
Z
0
g(x)dx−g(1)
2
; (11417.2)
3/2
Z
1
(g0(x))2dx≥3·2
2
3/2
Z
1
g(x)dx−g(1)
2
; (11417.3)
2
Z
3/2
(g0(x))2dx≥3·2
2
2
Z
3/2
g(x)dx−g(2)
2
; (11417.4)
3
Z
2
(g0(x))2dx≥3
3
Z
2
g(x)dx−g(2)
2
. (11417.5)
Proof of (11417.2). By partial integration,
1
Z
0
xg0(x)dx= xg(x)|1x=0
| {z }
=1g(1)−0g(0)=g(1)
−
1
Z
0
x0
|{z}=1
g(x)dx=g(1)−
1
Z
0
g(x)dx;
1
thus, by the integral Cauchy-Schwarz inequality,
1
Z
0
x2dx
| {z }
=13 1
Z
0
(g0(x))2dx≥
1
Z
0
xg0(x)dx
2
=
g(1)−
1
Z
0
g(x)dx
2
=
1
Z
0
g(x)dx−g(1)
2
.
This yields (11417.2).
Proof of (11417.3). By partial integration,
3/2
Z
1
(2x−3)g0(x)dx= (2x−3)g(x)|1x=0
| {z }
=(2·3/2−3)g(3/2)
−(2·1−3)g(1)
=0g(3/2)−(−1)g(1)
=g(1)
−
3/2
Z
1
(2x−3)0
| {z }
=2
g(x)dx=g(1)−2
3/2
Z
1
g(x)dx;
thus, by the integral Cauchy-Schwarz inequality,
3/2
Z
1
(2x−3)2dx
| {z }
=16
3/2
Z
1
(g0(x))2dx≥
3/2
Z
1
(2x−3)g0(x)dx
2
=
g(1)−2
3/2
Z
1
g(x)dx
2
=
2
3/2
Z
1
g(x)dx−g(1)
2
.
This yields (11417.3).
The inequalities (11417.4) and (11417.5) follow from (11417.3) and (11417.2), re- spectively, by substituting g1 for g, where the function g1 : [0,3] → R is defined by g1(x) = g(3−x) for all x∈[0,3].
Denote
A=
1
Z
0
g(x)dx−g(1), B = 2
3/2
Z
1
g(x)dx−g(1),
C = 2
2
Z
3/2
g(x)dx−g(2), D=
3
Z
2
g(x)dx−g(2).
2
Then,
A−B−C+D=
1
Z
0
g(x)dx−2
3/2
Z
1
g(x)dx−2
2
Z
3/2
g(x)dx+
3
Z
2
g(x)dx
=
1
Z
0
g(x)dx−2
3/2
Z
1
g(x)dx+
2
Z
3/2
g(x)dx
| {z }
=
2
R
1
g(x)dx
+
3
Z
2
g(x)dx
=
1
Z
0
g(x)dx+
2
Z
1
g(x)dx+
3
Z
2
g(x)dx
| {z }
=
3
R
0
g(x)dx
−3
2
Z
1
g(x)dx
| {z }
=0
=
3
Z
0
g(x)dx.
Thus,
3
Z
0
(g0(x))2dx=
1
Z
0
(g0(x))2dx
| {z }
≥3A2by (11417.2)
+
3/2
Z
1
(g0(x))2dx
| {z }
≥3·2B2by (11417.3)
+
2
Z
3/2
(g0(x))2dx
| {z }
≥3·2C2by (11417.4)
+
3
Z
2
(g0(x))2dx
| {z }
≥3D2by (11417.5)
≥3A2 + 3·2B2+ 3·2C2+ 3D2 = 3 A2 + 2B2+ 2C2+D2
≥3 A2+ (B+C)2+D2
since 2B2+ 2C2 = (B+C)2+ (B−C)2 ≥(B +C)2
= 12+ (−1)2+ 12
A2+ (B+C)2+D2
≥(1·A+ (−1)·(B+C) + 1·D)2 (by Cauchy-Schwarz)
= (A−B−C+D)2 =
3
Z
0
g(x)dx
2
,
proving (11417.1) and therefore solving the problem.
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