And finally, a question on number systems.?

Question

If a, b, c are three positive numbers such that a + b + c = 1, prove that ab + bc + ac ≤ 1/3

[HINT: Use a² + b² ≥ 2ab etc, & use square of trinomial]

Answer 1

a + b + c = 1
(a + b + c)² = 1
a² + b² + c² + 2(ab + bc + ac) = 1
a² + b² + c² = 1 - 2(ab + bc + ac)

a² + b² ≥ 2ab --(1)
b² + c² ≥ 2bc --(2)
a² + c² ≥ 2ac --(3)

(1)+(2)+(3),

2(a² + b² + c²) ≥ 2(ab + bc + ac)
a² + b² + c² ≥ ab + bc + ac
1 - 2(ab + bc + ac) ≥ ab + bc + ac
3(ab + bc + ac) ≤ 1
ab + bc + ac ≤ 1/3

Answer 2

ans.
1/r =1/a +1 /b + 1/c
.
1/r = bc/abc +ac/abc+ab/abc
.
1/r = ab+bc+ac /abc
ab+bc+ac = abc /r=abc/(3abc(a+b+c)) =<1/3