Question

if a2+b2+c2=1 then the value of ab+bc+ca= ?

Answer 1

HELLO

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

 Therefore,
 1 + 2(ab + bc + ca) ≥ 0_________ { by (a+b+c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)}

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)

so, ur answer is [-1/2, 1]

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THANKS ☺️

Answer 2

HEY Buddy......!! here is ur answer

We know that....(a+b+c)²=a²+b²+c²+2(ab+bc+ca)

Here, (a+b+c)² will be 0 for any real value of a, b and c

=> a²+b²+c²+2(ab+bc+ca) = 0

=> 1+2(ab+bc+ca) = 0

[ given that a²+b²+c² = 1 ]

=> ab+bc+ca = –1/2

I hope it will be helpful for you....!!

THANK YOU ✌️✌️

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