Prove that $n (3^{1/n} - 1) > 1$ for $n \in \mathbb{N}$!

Question

I've been trying to prove that $(n+1)^n \leq 3n^n$ and got down to prove that $n(3^{1/n}-1)>1$, but I can't go further from here, can anyone pls help?

Answer 1

$$3^{x}>e^{x}=1+\frac x {1!}+\frac {x^{2}} {2!}+...$$ so $3^{x}>1+x$ for all positive numbers $x$.

Take $x=\frac 1 n$.

Answer 2

Hint: Your inequality is equivalent to $$\left(1+\frac{1}{n}\right)^n\le 3$$, think about the Euler number $e$.

Answer 3

You can write $$n\left(3^\frac{1}{n}-1\right)=n \left(e^{\frac{1}{n}\log 3}-1 \right).$$Then your claim follows from $$e^y > 1+y,\quad y \neq 0$$ and the fact that $\log 3 > 1$.

Answer 4

Consider the sequence $s_n=(1+\frac 1n)^{(n+1)}$

Show that:

$(i)$ $s_n$ is decreasing, and

$(ii)$ $\forall n, s_n>t_n$, where $t_n = (1+\frac 1n)^n$

Edit:

Prove the following too:

$t_n$ is increasing. This helps you avoid too many menial calculations.

Answer 5

$f:x\mapsto 3^x$ is a convex function, hence the graph of $3^x$ lies above the tangent line at the origin.
This implies $3^x\geq 1+x\log 3$, and for any $x>0$ we have $3^x>1+x$ since $e<3$.
The claim follows by evaluating the inequality $3^x>1+x$ at $x=\frac{1}{n}$.