Question

If a b c are distinct positive real numbers and a^2+b^2+c^2=1 then ab+bc+ca id

Answer 1

Answer:

(a + b + c)² = a² + b² + c² + 2(ab + bc + ca)

2(ab + bc + ca) = (a + b + c)² – (a² + b² + c²)

(ab + bc + ca) = [(a + b + c)² – 1]/2

Since a2 + b2 + c2 = 1, all a, b and c are less than 1.

So, (ab + bc + ca) must be less than 1.

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Answer 2

ab + bc + ca lies in $(-\dfrac{1}{2},1)$

Step-by-step explanation:

Given,

The distinct positive real numbers = a, b and c

$a^2+b^2+c^2=1$

To find, the value of ab + bc + ca = ?

We know that,

$(a+b+c)^2>0$

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

⇒ $1+2(ab+bc+ca)>0$

⇒ $ab+bc+ca>-\dfrac{1}{2}$     ......(1)

Also,

$(a-b)^{2}+(b-c)^{2}+(c-a)^{2}>0$

⇒ $a^2+b^2-2ab+b^2+c^2-2bc+c^2+a^2-2ca>0$

⇒ $2(a^2+b^2+c^2)-2(ab+bc+ca)>0$

⇒ 1-(ab+bc+ca)>0[/tex]

⇒ $-(ab+bc+ca)>-1$

⇒ $ab+bc+ca>1$       .....(2)

From (1) and (2), we get

∴ ab + bc + ca lies in $(-\dfrac{1}{2},1)$