Question

find the minimum value of x² + y² + z² when ax + by + cz = p
​

Answer 1

Using Vectors or Couchy Schwartz Inequality (Both are Equivalent here)

Lets take two vectors

⃗=⟨,,⟩
A
→
=
⟨
a
,
b
,
c
⟩


⃗=⟨,,⟩
X
→
=
⟨
x
,
y
,
z
⟩


Now the expression can be written as

++=⃗⋅⃗
a
x
+
b
y
+
c
z
=
A
→
⋅
X
→


hence

⃗⋅⃗=
A
→
⋅
X
→
=
p


Now using another definition of dot product we have

⃗⋅⃗=2+2+2‾‾‾‾‾‾‾‾‾‾‾‾√2+2+2‾‾‾‾‾‾‾‾‾‾‾‾√cos()=
A
→
⋅
X
→
=
a
2
+
b
2
+
c
2
x
2
+
y
2
+
z
2
cos
⁡
(
θ
)
=
p


where
θ
is the angle between the two vectors defined above ⃗⃗
A
→
a
n
d
X
→


Now since

cos()≤1
cos
⁡
(
θ
)
≤
1


Or 2+2+2‾‾‾‾‾‾‾‾‾‾‾‾√2+2+2‾‾‾‾‾‾‾‾‾‾‾‾√≤1
p
a
2
+
b
2
+
c
2
x
2
+
y
2
+
z
2
≤
1


22+2+2≤2+2+2
p
2
a
2
+
b
2
+
c
2
≤
x
2
+
y
2
+
z
2


So The minimum value of the expression is

22+2+2
p
2
a
2
+
b
2
+
c
2