Shaun from Nottingham High School sent this
solution.
(i) By calculation, we have
|
P = |
æ
ç
è
|
ta+b+1
a+b+1
|
ö
÷
ø
|
2
|
= |
t2(a+b+1)
(a+b+1)2
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, |
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and
|
Q = |
æ
ç
è
|
t2a+1
2a+1
|
ö
÷
ø
|
|
æ
ç
è
|
t2b+1
2b+1
|
ö
÷
ø
|
= |
t2(a+b+1)
(2a+1)(2b+1)
|
. |
|
Thus we must decide which of the two expressions
is the smaller.
Since
|
(a+b+1)2-(2a+1)(2b+1) = (a-b)2 ³ 0 |
|
we see that P £ Q.
(ii) Given the inequality
|
|
ó
õ
|
t
0
|
[f(x)+lg(x)]2 dx ³ 0, |
|
we have
|
l2 |
ó
õ
|
t
0
|
g2(x)dx+2l |
ó
õ
|
t
0
|
g(x)f(x)dx + |
ó
õ
|
t
0
|
f2(x)dx ³ 0 |
|
The discriminant of this quadratic in
l must be less than or equal to 0 since the quadratic is
never negative:
|
4[ |
ó
õ
|
t
0
|
g(x)f(x)dx]2-4 |
ó
õ
|
t
0
|
g2(x)dx |
ó
õ
|
t
0
|
f2(x)dx £ 0. |
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This gives the required inequality which is known as the Cauchy
Schwarz inequality:
|
|
æ
è
|
ó
õ
|
t
0
|
f(x)g(x) dx |
ö
ø
|
2
|
£ |
æ
è
|
ó
õ
|
t
0
|
f(x)2 dx |
ö
ø
|
|
æ
è
|
ó
õ
|
t
0
|
g(x)2 dx |
ö
ø
|
. |
|