Newton's Beautiful Problem
⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ x + y + z = 6 x 2 + y 2 + z 2 = 1 2 x 3 + y 3 + z 3 = 2 4
Find x ! × y ! × z ! .
The answer is 8.
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3 solutions
In order to apply CS, why must these terms be real?
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CS works only on inner products which are scalars, real values.
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My point is that you are making the assumption that x , y , z are real values.
Note: CS works on inner products, even complex inner product, where < u , v > = u v . So, it doesn't necessarily need to be real values, but just evaluated correctly.
x
y
+
y
z
+
x
z
=
2
(
x
+
y
+
z
)
2
−
(
x
2
+
y
2
+
z
2
)
=
2
3
6
−
1
2
=
1
2
=
x
2
+
y
2
+
z
2
∴
x
2
+
y
2
+
z
2
−
x
y
−
y
z
−
x
z
=
0
∴
2
(
x
−
y
)
2
+
(
y
−
z
)
2
+
(
x
−
z
)
2
=
0
→
x
=
y
=
z
Using equation 1,
3
x
=
6
,
→
x
=
y
=
z
=
2
x
!
y
!
z
!
=
2
3
=
8
Newton's sums method:
x + y + z = 6 x 2 + y 2 + z 2 = ( x + y + z ) 2 − 2 ( x y + x z + y z ) 1 2 = 6 2 − 2 ( x y + x z + y z ) ⟹ x y + x z + y z = 1 2 x 3 + y 3 + z 3 = ( x + y + z ) ( x 2 + y 2 + z 2 ) − ( x y + x z + y z ) ( x + y + z ) + 3 x y z 2 4 = 6 ( 1 2 ) − 1 2 ( 6 ) + 3 x y z ⟹ x y z = 8
Use this to form a cubic equation:
a 3 − 6 a 2 + 1 2 a − 8 = 0 a 3 − 3 ( 2 ) a 2 + 3 ( 2 2 ) a − 2 3 = 0 ( a − 2 ) 3 = 0 ⟹ a = 2
Therefore, we know that x = y = z = 2
⟹ x ! × y ! × z ! = 2 ! × 2 ! × 2 ! = 8
Using Cauchy-Schwartz inequality , we have:
( x + y + z ) 2 3 6 ≤ 3 ( x 2 + y 2 + z 2 ) = 3 6
We note that LHS=RHS, meaning equality occurs and x = y = z = 2 , ⟹ x 2 + y 2 + z 2 = 1 2 and x 3 + y 3 + z 3 = 2 4 . Therefore, x ! y ! z ! = 8 .