Prove that $2^{n}>1+n\sqrt{2^{\left(n-1\right)}}, \ \ \forall n>2$.

this is a question from the concept of $$AM\ge GM\ge HM$$ , how do i know which number to select for applying the inequality, please help!

• Hint: use $$2^n-1=1+2+2^2+...2^{n-1}\ge ?$$ Commented Feb 20, 2021 at 2:56
• @Albus Dumbledore Thanks a lot! I got it, i was able to solve it! But isn't a bit hard to find how to start these questions? I mean, how did you figure out how to use this? Commented Feb 20, 2021 at 3:02
• John Wick not really, notice that $$\sqrt{2^{n-1}}=\sqrt[n]{2^{\frac{n(n-1)}{2}}}$$ gave the clue Commented Feb 20, 2021 at 3:04

This is the special case of $$q=2$$ from the following inequality.

$$q^k \ge 1+k(q-1)\sqrt{q^{k-1}}$$

To prove this, first remember that AM>= GM, so we want the 'add' on the left and the 'multiply' on the right.

We have:

$$\frac{q^k-1}{q-1}=1+q+q^2+\dots+q^{k-1}$$

and also

$$\prod_\limits{i=0}^{k-1} q^i = q^{\frac{k(k-1)}{2}}$$

AMGM tells us that:

$$\frac{1+q+q^2+\dots+q^{k-1}}{k} \ge \sqrt[k]{q^{\frac{k(k-1)}{2}}}$$

$$\frac{q^k-1}{k(q-1)} \ge \sqrt[k]{q^{\frac{k(k-1)}{2}}}$$

$$\frac{q^k-1}{k(q-1)} \ge \sqrt{q^{k-1}}$$

$$q^k \ge 1+k(q-1)\sqrt{q^{k-1}}$$

Note that if $$q<1$$, then $$(q-1)$$ is negative, and this reverses the inequality.