DrAymeinstein wrote: Let $a>1$ be a real number. Prove that for all $n\in\mathbb{N}*$ that : $\frac{a^n-1}{n}\ge \sqrt{a}^{n+1}-\sqrt{a}^{n-1}$ We need to prove that $f(a)\geq0,$ where $$f(a)=a^{2n}-na^{n+1}+na^{n-1}-1.$$Now, by AM-GM $$f'(a)=2na^{2n-1}-n(n+1)a^n+n(n-1)a^{n-2}=na^{n-2}\left(2a^{n+1}-(n+1)a^2+n-1\right)\geq0,$$which says $f(a)\geq f(1)=0.$

$$a^n-1=(a-1)(a^{n-1}+a^{n-2}+\cdots+a+1)\geq (a-1)\cdot n\sqrt{a^{n-1}}=n\left(\sqrt{a^{n+1}}-\sqrt{a^{n-1}}\right)$$

same solution as @ehuseyinyigit

solution

motivation