Question

If a
2
+ b
2
+ c
2
= 14, then ab + bc + ca is always
greater than or equal to

Answer 1

(a+b+c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)

 

(a+b+c)^2 ≥ 0 for any real values of a, b, c

 

Therefore,  

a^2 + b^2 + c^2 + 2(ab + bc + ca) ≥ 0

 

Given that a^2 + b^2 + c^2 = 1

Therefore,

1 + 2(ab + bc + ca) ≥ 0

 

ab + bc + ca ≥ -1/2

(a-b)^2 + (b-c)^2 + (c-a)^2 ≥ 0

2 [ a^2 + b^2 + c^2 - ab - bc - ca ] ≥ 0

2 [ 1 - (ab + bc + ca)] ≥ 0

Therefore, 1 ≥ (ab + bc + ca)

Hence the answer is [-1/2, 1

Answer 2

ab+bc+ca is always greater than or equal to one