Question

Ifa2+b2+c2=14;then ab+bc+ca is always greater than or equal to

Answer 1

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

$a^2+b^2+c^2+2ab+2bc+2ca\geq 0$

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

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

$(ab+bc+ca)\geq -7$

$(ab+bc+ca)\leq 7$

Now, consider $(a-b)^2+(b-c)^2+(c-a)^2\geq 0$

$a^2+b^2-2ab+b^2+c^2-2bc+c^2+a^2-2ac\geq 0$

$2a^2+2b^2+2c^2-2ab-2bc-2ac\geq 0$

$2(a^2+b^2+c^2)-2(ab+bc+ac)\geq 0$

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

$28-2(ab+bc+ac)\geq 0$

$-2(ab+bc+ac)\geq -28$

$(ab+bc+ac)\geq 14$

Therefore, $(ab+bc+ac)\geq 14$.

Answer 2

Thank you for asking this question, here is your answer.

a² + b² + c² = 14

In this situation we are aware of the fact that the square of every number is greater than or equal to 0.

now we will take the square of (a+b+c)

(a+b+c)²≥ 0

a² + b² + c²  + 2(ab + bc + ca) ≥ 0

14 +2 (ab + bc + ca) ≥ 0

2 (ab + bc + ca) ≥ - 14

ab  + bc + ca ≥ -14/2

ab + bc + ca ≥ - 7

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