Prove by induction that $\sqrt[n+1]{n+1}<\sqrt[n]{n}$ for $n\ge 3$

Question

I've been dealing with this problem for almost 2 hours now, with hardly any progress. I'm to prove the following inequality using induction

$$\sqrt[n+1]{n+1}<\sqrt[n]{n}$$ where $n≥3$, $n∈\mathbb{N}$. How can this be done?

Answer 1

HINT: your inequality is equivalent to $$\left(1+\frac{1}{n}\right)^n<n$$

Answer 2

I have finally managed it, my solution goes as follows: First we raise both side of the inequality to the power $n(n+1)$, and so we obtain: $$(n+1)^{n}<n^{n+1}$$ Now we multiply both side by $\frac{(n+1)^{n+2}}{n^{n+1}}$. Now we get $$\frac{(n+1)^{2n+2}}{n^{n+1}}<(n+1)^{n+2}$$ After some work on the left side of this inequality we obtain $$\frac{(n+1)^{2n+2}}{n^{n+1}}=\frac{(n+1)^{2(n+1)}}{n^{n+1}}=\left(\frac{n^{2}+2n+1}{n}\right)^{n+1}=\left(n+2+\frac{1}{n}\right)^{n+1}>(n+2)^{n+1}$$ Now all that's left is to raise both sides of the inequality to the power $\dfrac{1}{(n+1)(n+2)}$, and so we get $$\sqrt[n+2]{n+2}<\sqrt[n+1]{n+1}$$ QED