=
h
u
,
u
i  h
u
,
v
i  h
v
,
u
i
+
h
v
,
v
i
=

u

2
 h
u
,
v
i  h
v
,
u
i
+

v

2
.
Adding these two equations yields

u
+
v

2
+

u

v

2
= 2

u

2
+ 2

v

2
.
3. (2 marks) Let
V
be an inner product space over
R
. Suppose
u
,
v
∈
V
have the
same norm. Prove that
u
+
v
and
u

v
are orthogonal.
Solution
: We have
h
u
+
v
,
u

v
i
=
h
u
,
u

v
i
+
h
v
,
u

v
i
=
h
u
,
u
i  h
u
,
v
i
+
h
v
,
u
i  h
v
,
v
i
=

u

2
 h
u
,
v
i
+
h
u
,
v
i  
v

2
= 0
(we used the fact that

u

=

v

, and that since the field of scalars is
R
, inner
product space axiom P5 yields
h
u
,
v
i
=
h
v
,
u
i
).
4. (2 marks) Let
x
1
, ..., x
n
∈
R
. Prove that
(
x
1
+
x
2
+
...
+
x
n
)
2
≤
n
(
x
2
1
+
x
2
2
+
...
+
x
2
n
)
(hint:
use CauchySchwarz).
Deduce that the square of an average of real
numbers is less than or equal to the average of the squares.
Solution
: We use (the square of) the CauchySchwarz inequality in
R
n
with
u
= (1
,
1
, ...,
1) and
v
= (
x
1
, x
2
, ..., x
n
). This yields
(
x
1
+
x
2
+
...
+
x
n
)
2
=
h
u
,
v
i
2
≤ 
u

2

v

2
=
n
(
x
2
1
+
x
2
2
+
...
+
x
2
n
)
.
Now, starting with the square of the average of
x
1
, x
2
, ..., x
n
, we have
x
1
+
x
2
+
...
+
x
n
n
2
=
(
x
1
+
x
2
+
...
+
x
n
)
2
n
2
≤
n
(
x
2
1
+
x
2
2
+
...
+
x
2
n
)
n
2
=
x
2
1
+
x
2
2
+
...
+
x
2
n
n
,
which is the average of the squares.
5. (2 marks) Let
a, b, c >
0 be real numbers. Prove that
abc
(
a
+
b
+
c
)
≤
a
3
b
+
b
3
c
+
c
3
a
(hint: use CauchySchwarz on
a
√
c
,
b
√
a
,
c
√
b
and (
√
c,
√
a,
√
b
)).