Prove $\left(2017^{2018}+2017^{2017}\right)^{2018}>\left(2018^{2017}+2017^{2017}\right)^{2017}$ Can someone give me a hint for this problem (for secondary student): 

Prove that: $$\left(2017^{2018}+2017^{2017}\right)^{2018}>\left(2018^{2017}+2017^{2017}\right)^{2017}$$ 

P/s: I've thinking about using the fact that $n^{n+1}>(n+1)^n$ for $n$ is a natural number, but to prove this I've to use the derivative.
 A: Hint:
$$
\frac{2017^{2018}}{2017^{2017}}=2017
$$
but
$$
\begin{align}
\frac{2018^{2017}}{2017^{2017}}
&=\left(1+\frac1{2017}\right)^{2017}\\
&\le\left(1+\frac1{2017}\right)^{2018}\\
&\le\left(1+\frac1{2016}\right)^{2017}\\
&\qquad\ \ \vdots\\
&\le\left(1+\frac11\right)^2\\[6pt]
&=4
\end{align}
$$
To show the last series of inequalities, we can use Bernoulli's Inequality, which has a very simple inductive proof:
$$
\begin{align}
\frac{\left(1+\frac1{n-1}\right)^n}{\left(1+\frac1n\right)^{n+1}}
&=\frac{n-1}n\left(1+\frac1{n^2-1}\right)^{n+1}\\
&\ge\frac{n-1}n\left(1+\frac1{n-1}\right)\\[9pt]
&=1
\end{align}
$$
A: As the OP indicates, the desired inequality follows easily from the general inequality $(n+1)^n\lt n^{n+1}$, which holds for $n\ge3$ (but not for $n=1$ or $2$).  Here's the essential step in an easy induction proof.
Rewriting the inequality to prove as $\left( n+1\over n\right)^n\lt n$, let's assume by induction that we have
$$\left( n\over n-1\right)^{n-1}\lt n-1$$
Then, from ${n+1\over n}\lt {n\over n-1}$ (since $(n+1)(n-1)=n^2-1\lt n^2$), we have
$$\left( n+1\over n\right)^n\lt\left( n\over n-1\right)^n=\left( n\over n-1\right)\left( n\over n-1\right)^{n-1}\lt\left( n\over n-1\right)(n-1)=n$$
A: $n^{n+1}>(n+1)^n$ for $n>3$
Let $n=2017$, then your inequality becomes
$$(n^{n+1}+n^n)^{n+1}>((n+1)^n+n^n)^n$$
Proof: $(n^{n+1}+n^n)^{n+1}>((n+1)^n+n^n)^{n+1}$ since $a>b$ implies $a^c>b^c$ for positive $a,b$ and integer $c$ and $((n+1)^n+n^n)^{n+1}>((n+1)^n+n^n)^n$ since $(n+1)^n+n^n>1$
A: The numbers they use in context-like problems are usually irrelevant
Focus on $$\left(n^{n+1}+n^n\right)^{n+1}>\left(n^n+(n+1)^n\right)^n$$ 
A key point is to prove that
$n^{n+1}>(n+1)^n;\;\forall n\ge 3\quad$ 
If you divide both sides by $n^n$ you get 
$$n>\left(\frac{1}{n}+1\right)^n$$ 
which is true for $n\ge 3$ because all the $(n+1)$ terms of the binomial power in the RHS are (much) less than or equal to $1$ and their sum is less than $n$. For instance for $n=4$ it is $\frac{1}{4^4}+\frac{4}{4^3}+\frac{6}{4^2}+\frac{4}{4}+1$ which is way smaller than $4$. Namely it's about $2.44$. 
It is not a rigorous proof, but for a $9$th grader I hope it is understandable
Hope this can help
A: Hint: it is enough to show $x^{1/x}$ is decreasing for $x>e$.
A: Let $2017=n$.
Thus, we need to prove that
$$\left(n^{n+1}+n^n\right)^{n+1}>\left((n+1)^n+n^n\right)^n$$ or
$$(n+1)^{n+1}n^{n(n+1)}>\left((n+1)^n+n^n\right)^n$$ or
$$(n+1)^{1+\frac{1}{n}}n^{n+1}>(n+1)^n+n^n$$ or
$$n^n\left(n(n+1)^{1+\frac{1}{n}}-1\right)>(n+1)^n$$ or
$$n(n+1)^{1+\frac{1}{n}}-1>\left(1+\frac{1}{n}\right)^n,$$
which is obvious because $$\left(1+\frac{1}{n}\right)^n<3.$$
It's enough to prove that $(n+1)^n<n^{n+1}$ for all $n\geq3$, which we can prove by induction.
We need to prove that 
$$\left(1+\frac{1}{n}\right)^n<n$$ and since
$$\left(1+\frac{1}{n+1}\right)^{n+1}<\left(1+\frac{1}{n}\right)^{n+1}<n\left(1+\frac{1}{n}\right)=n+1,$$
it remains to check a base induction, which is obvious.
