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If n is a positive integer then how can I prove that $$2^n>1+n \sqrt{2^{n-1}}$$ .Any hint may help.My textbook mentions this problem in category of A.M. ,G.M. , H.M. inequalities.So please give hint according to that.Im also comfortable with using Weighted means and Cauchy Schwartz inequality.

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  • $\begingroup$ The inequality is false for $n=1.$ Is it $n>1?$ $\endgroup$ – mfl Nov 10 '18 at 12:19
  • $\begingroup$ @mfl Either that or the sign should be changed to $\geq$. $\endgroup$ – AryanSonwatikar Nov 17 '18 at 16:06
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An answer using that the arithmetic mean of $n$ positive real numbers is largerthan or equal to their geometric mean: we have $2^n-1=\sum_{m=0}^{n-1}2^m = n \left[\frac{1}{n}\sum_{m=0}^{n-1}2^m\right] $ which is $n$ times the arithmetic mean of those powers of $2$, whose geometric mean is
$$(2^02^1\ldots2^{n-1})^{1/n}=\left(2^{\sum_{m=0}^{n-1}m}\right)^{1/n}=\left(2^{n(n-1)/2}\right)^{1/n}=\sqrt{2}^{n-1}.$$ Thus $$2^n-1\geq n \sqrt{2}^{n-1}$$ and the statement follows, with equality for $n=1$ as pointed out in the comment by mfl.

edit: the first equality results, for instance, from the geometric sum $\sum_{m=0}^{n-1}x^m = \frac{1-x^n}{1-x}$ for $x=2$.

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$$ \\1+n\sqrt{2^{n-1}}\le n\sqrt{2^{n-1}}+n\sqrt{2^{n-1}}=2n\sqrt{2^{n-1}}=n2^{(n+1)/2}\le2^{(n-1)/2}2^{(n+1)/2}=2^n $$ $2^{(n-1)/2}\ge n$ it is prove induction for n. $n=1$ then $1\ge1$. Let for $n=m\;$ $2^{(m-1)/2}\ge m=>2^{m/2}=2^{(m-1)/2+1/2}\ge m\cdot 2^{1/2}=m+m(2^{1/2}-1)>m+1\;(n>1=>2^{(n-1)/2}>n)$

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