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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. |
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Answer» 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. |
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