<s>[INST] Prove that $n(1+n)^{\frac{1}{n}} < n+H_n$ for every $n \geq 2$. 
For every positive integer $n$ set $H_n = \dfrac{1}{1}+\dfrac{1}{2}+\cdots+\dfrac{1}{n}$. Prove that $n(1+n)^{\frac{1}{n}} < n+H_n$ for every $n \geq 2$.

Attempt
I will prove this result by induction. The base case holds since for $n = 2$ we have  $(1+2)^{\frac{1}{2}} < 2 + H_2 = 3.5$. Now assume that $k(1+k)^{\frac{1}{k}} < k+H_k$ holds for some $k$. We need to show that $(k+1)(2+k)^{\frac{1}{k+1}} < 1+k+H_{k+1}$. How do I show this? Also, is induction the best way to prove this? 
 [/INST] Notice that $$n+1 = \frac{2}{1}\frac{3}{2} \cdots \frac{n}{n-1}\frac{n+1}{n}   $$
Hence by AM-GM we get
$$ \left (n+1\right )^{\frac{1}{n}} < \frac{ \frac{2}{1}+\frac{3}{2}+ \cdots \frac{n}{n-1}+\frac{n+1}{n}     }{n} $$
But observe that 
$$\frac{ \frac{2}{1}+\frac{3}{2}+ \cdots \frac{n}{n-1}+\frac{n+1}{n}     }{n}= \frac{n + H_n}{n}.$$
And we are done.
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