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[Post New]25 Apr 2008 18:40:13 IST
    Subject: Re:Prove that for allintegers n2, 2n-1 (3n + 4n) >7n
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hsbhatt (4990)

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Olaaa!! Perrrfect answer. 940  [1085 rates]
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There is an inequality that is known as the Power Mean Inequality. The version we will use here is:\left (\frac{a^n+b^n}{2} \right)^\frac{1}{n} \ge \frac{a+b}{2}


Here a =3, b =4and you get 2^{n-1} (3^n+4^n) \ge (3+4)^n = 7^n


PS: Can u try using Chebyshev's Inequality for this one?
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