Question

If a2 + b2 + c2 = 14, then ab + bc + ca is always
greater than or equal to

Answer 1

It is given that ,a²+ b²+c²=14

Now, as we know

→(a+b+c)²=a²+b²+c²+2 a b+ 2 b c+ 2 c a

 →(a+b+c)²= 14+2 (a b+  b c+  c a) [ ∴a²+ b²+c²=14 ]

as we know that square of any real number is always greater than zero.

So,(a+b+c)²≥0

→14+2 (a b+  b c+  c a )≥ 0 [∴(a+b+c)²= 14+2 (a b+  b c+  c a) ]

→ 2 (a b+  b c+  c a )≥ 0 - 14

→  2 (a b+  b c+  c a )≥  - 14

Dividing both side by 2,we get

→ a b+  b c+  c a ≥  - 7, which is the desired result.

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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