Question #87182
Prove that 2^n >1+n√[2^(n-1)] for n>2, using the Inequalities
1
Expert's answer
2019-04-03T11:06:18-0400

2n>1+n2n12^n >1+n\sqrt{2^{n-1}} for n>2n>2

2n1n2n1>02^n-1-n\sqrt{2^{n-1}}>0

taking derivative of the left part

2nln(2)(n2n1ln(2))/22n12^n*ln(2)-(n*\sqrt{2^{n-1}}*ln(2))/2-\sqrt{2^{n-1}}

by looking up/making a graph of this function you can see that for n>2 its always positive. It means that that left part will always grow for n>2, so now we only need to check for the inequality to be true for n=2,


2212221>02^2-1-2\sqrt{2^{2-1}}>0

322>03-2*\sqrt2>0

2\sqrt2 is approximately 1.414 so the inequality is true for n=2 and any n>2


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