Question

If a, b, c are all positive, a + b + c = 1 and (1 – a) (1 – b) (1 – c) ≥ K(abc), then find the value of K.

Answer 1

Given, a + b + c = 1, so that a + b = 1 – c, b + c = 1 – a, c + a = 1 – b 

Now, AM > GM ⇒ (a + b) > 2\(\sqrt{ab}\) ⇒ (1 – c) > 2\(\sqrt{ab}\)       …(i) 

Similarly, (b + c) > 2\(\sqrt{bc}\)  ⇒ (1 - a) > 2\(\sqrt{ab}\)            …(ii) 

(c + a) > 2\(\sqrt{ca}\)  ⇒ (1 - b) > 2\(\sqrt{ca}\)                            …(iii) 

Multiplying (i), (ii) and (iii), we get 

⇒ (1 – a) (1 – b) (1 – c) > 8 \(\sqrt{ab}\) \(\sqrt{bc}\) \(\sqrt{ca}\)

⇒ (1 – a) (1 – b) (1 – c) > 8 abc ⇒ K = 8.