Greatest value
a 2 + b 2 = 7 and a 3 + b 3 = 1 0 , then the greatest value of a + b can be
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2 solutions
Moderator note:
You still have to show that these values are attainable. Currently, all that you have is a necessary condition. Does there exist a + b = 4 that satisfy the conditions?
Nice solution! But don't you think that the Cauchy-Schwarz Inequality would have lessened our labours a lot?
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Not quite. This approach provides a much more restrictive condition than CS, which gives you more insight into the problem, namely that the value of a + b is pretty limited.
Probably the reason for several option suggest the approach below
at this level.
7
=
a
2
+
b
2
=
(
a
+
b
)
2
−
2
a
b
.
∴
a
b
=
2
(
a
+
b
)
2
−
7
.
1
0
=
a
3
+
b
3
=
(
a
+
b
)
3
−
3
a
b
(
a
+
b
)
=
(
a
+
b
)
{
(
a
+
b
)
2
−
3
a
b
}
.
∴
a
b
=
3
(
a
+
b
)
2
−
a
+
b
1
0
.
∴
2
(
a
+
b
)
2
−
7
=
3
(
a
+
b
)
2
−
a
+
b
1
0
.
⟹
(
a
+
b
)
{
2
1
−
(
a
+
b
)
2
}
=
2
0
.
L
.
H
.
S
.
w
i
l
l
g
i
v
e
f
r
a
c
t
i
o
n
f
o
r
a
+
b
=
2
9
.
∴
n
o
t
a
s
o
l
u
t
i
o
n
.
a
+
b
=
4
,
L
.
H
.
S
.
=
4
∗
(
2
1
−
1
6
)
=
2
0
=
R
.
H
.
S
.
a
+
b
=
4
.
a+b=5, R.H.S. >0. So~(a+b)^2>21, no solution > 4.
At this level
the wording "greatest value" only miss leads. If that is the intention it is OK.
a + b = x ( s a y ) , then x 2 = a 2 + b 2 + 2 a b , x 3 = a 3 + b 3 + 3 a b ( a + b ) , Eliminating a b , we get x 3 − 2 1 x + 2 0 , So, x = 1 or 4 or − 5