Can $k>1$ Prove This Inequality?

[anemone]

Prove that for all integers $k>1$:

$\left(\dfrac{1+(k+1)^{k+1}}{k+2}\right)^{k-1}>\left(\dfrac{1+k^k}{k+1}\right)^{k}$

 

[anemone]

Hint:

Spoiler

Compare the value of $k^{k(k-1)}$ with both

1. $\left(\dfrac{1+(k+1)^{k+1}}{k+2}\right)^{k-1}$ and

2. $\left(\dfrac{1+k^k}{k+1}\right)^{k}$.

 

[anemone]

Solution of other:

Spoiler

Note that $\left(\dfrac{1+(k+1)^{k+1}}{k+2}\right)^{k-1}>k^{k(k-1)}>\left(\dfrac{1+k^k}{k+1}\right)^{k}---(1)$

Taking roots of order $k-1$ and $k$ respectively, we rewrite the left and right inequalities of (1) into their equivalent forms:

$\dfrac{(k+1)^{k+1}+1}{k+2}>k^k$ and $k^{k-1}>\dfrac{k^k+1}{k+1}---(2)$

Then second inequality is immediate as:

$(k+1)k^{k-1}=k^k+k^{k-1}>k^k+1$ for $k>1$.

For the first one for $k>1$, apply the Binomial Theorem to get

$\begin{align*}(k+1)^{k+1}+1&=(k^{k+1}+(k+1)k^k+\cdots+1)+1 \\&> k^{k+1}+(k+1)k^k\\&>k^{k+1}+2k^k=k^k(k+2)\end{align*}$

Hence both sides of the inequalities are proved and the result follows.