Proving the inequality $2^n ≤2^{n+1}−2^{n−1}−1$

Question

I've tried multiplying each side by $2^n$, factoring out a 2, but I find myself going in a circle. What's the best way to solve this?

$2^n ≤2^{n+1}−2^{n−1}−1$. For all integers n

Answer 1

Hint. I guess that $n\geq 1$. Let $x=2^{n-1}$ then show that $2x\leq 4x-x-1$.

Answer 2

$$(3k+2)/(2)\leq 2k$$. for all $k>0$ Check by PMI(principal of mathematical induction) Now,$ (3k/2) + 1 \leq 2k$ $ k + k/2 + 1 \leq2k$

Now take $k= 2^n$

$2^n +2^{n-1} +1 \leq2*2^n$ i.e $2^n \leq 2^{n+1} - 2^{n-1} - 1$

Answer 3

For $n$ integer $$2^{n+1}-2^{n-1}-1=4\cdot2^{n-1}-2^{n-1}-1=3\cdot 2^{n-1}-1=2^n+2^{n-1}-1$$ Then $2^{n+1}-2^{n-1}-1\ge 2^{n}\iff2^{n-1}\ge 1\iff n\ge 1$.

Answer 4

$$ 2^n = 2\cdot 2^{n-1} \le 3\cdot 2^{n-1}-1 =2^{n+1}-2^{n-1}-1. $$

Answer 5

As abiessu wrote, since $2^{n+1}=2\cdot 2^n$, this is $0\le 2^n-2^{n-1}-1 =2^{n-1}-1 $ which is true for $n \ge 1$.

The inequality is strict for $n \ge 2$.

Answer 6

It's$$2\cdot2^{n-1}\leq4\cdot2^{n-1}-2^{n-1}-1$$ or $$2^{n-1}\geq1.$$ Done!