<s>[INST] Prove that for $n\in N^*$ and $a,x \geq 0$ : $(a+x)^{n+1} \geq \frac{(n+1)^{n+1}}{n^n}a^nx$ 
Prove that for $n\in N^*$ and $a,x \geq 0$ : $(a+x)^{n+1} \geq \frac{(n+1)^{n+1}}{n^n}a^nx$

My proof :
$$(a+x)^{n+1} = \bigg(n\frac{a}{n} + x\bigg)^{n+1} = \bigg( \frac{a}{n} + \frac{a}{n} + \dots +\frac{a}{n} +x \bigg)^{n+1} $$
$$\geq$$
$$\bigg[ (n+1)\bigg(\frac{a}{n}\cdot\frac{a}{n}\cdot \dots \cdot \frac{a}{n}\cdot x\bigg)^{\frac{1}{n+1}} \bigg]^{n+1}$$
$$= $$
$$\bigg\{ (n+1) \bigg[ \bigg(\frac{a}{n}\bigg)^nx\bigg]^{\frac{1}{n+1}} \bigg\}^{n+1}$$ 
$$=$$
$$\frac{(n+1)^{n+1}}{n^n}a^nx$$
Wanted to ask if my approach seems correct and smooth enough, or if there's a better/easier way to prove it.
 [/INST] use the AM-GM inequality, then
$$\frac{\frac{a}{n}+\frac{a}{n}+...+\frac{a}{n}+x}{n+1}\geq \sqrt[n+1]{\left(\frac{a}{n}\right)^n\cdot x}$$
</s>