Question

Evaluate (ab+bc+ca+a²+b²–c²+1)÷ 2a for a = 1,b=2 and c=3.​

Answer 1

Answer:

we know that

(a+b+c)

2

=a

2

+b

2

+c

2

+2ab+2bc+2ca and

(a+b+c)

2

≥0 for any real a,b,c

Given, a

2

+b

2

+c

2

=1

Therefore, 1+2(ab+bc+ca)≥0

(ab+bc+ca)≥−

2

1

Since, A.M.≥G.M.

⟹

2

a+b

≥

ab

⟹a+b≥2

ab

Assume a=a

2

and b=b

2

⟹a

2

+b

2

≥2ab ------(1)

similarly,

b

2

+c

2

≥2bc ------(2)

c

2

+a

2

≥2ac ------(3)

adding (1), (2) and (3) we get

a

2

+b

2

+c

2

≥ab+bc+ca

Since, a

2

+b

2

+c

2

=1

(ab+bc+ca)≤1

Therefore, ab+bc+ca lies in the interval [−

2

1

,1]