Question

If a square + b square + c square is equal to 14 then find ab+bc+ca

Answer 1

To solve this question, you must know the identity

(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]

Hav a good time :)

Answer 2

If a square + b square + c square is equal to 14 then  $ab+bc+ca \geq -7$

Solution:

Given that,

$a^2 + b^2 + c^2 = 14$

To find : ab + bc + ca

We know that,

$(a+b+c)^2 = a^2 + b^2 + c^2 + 2 ( a b + b c + c a)$

$(a + b +c)^2 \geq 0$

Since, square of any real number is always ≥ 0

Therefore,

$14 + 2(ab+bc+ca)\geq 0\\\\2(ab+bc+ca)\geq - 14\\\\ab+bc+ca \geq -7$

Thus found

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