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If a, b, c are positive real numbers such that a + b + c = p, then \(\frac1a+\frac1b+\frac1c\) is greater than(a) 16 (b) 9p (c) \(\frac9p\)(d) \(\frac{16}{p}\) |
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Answer» (c) \(\frac9p\) The AM of mth powers of n positive numbers is greater than the mth power of their AM if m < 0 or m > 1. So for m = – 1, \(\frac{a^{-1}+b^{-1}+c^{-1}}{3}\) > \(\big(\frac{a+b+c}{3}\big)^{-1}\) ⇒ \(\frac13\)\(\big(\frac1a+\frac1b+\frac1c\big)\) > \(\big(\frac{p}{3}\big)^{-1}\) (Given: a + b + c ≡ p) ⇒\(\big(\frac1a+\frac1b+\frac1c\big)\) > \(\frac9p\) |
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