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