Let $a>1$ be a real number. Prove that for all $n\in\mathbb{N}*$ that : $\frac{a^n-1}{n}\ge \sqrt{a}^{n+1}-\sqrt{a}^{n-1}$
Problem
Source: Moroccan TST 2019 P2
Tags: inequalities, morocco
arqady
23.08.2024 19:39
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 $f(a)\geq f(1)=0.$
ehuseyinyigit
23.08.2024 20:55
$$a^n-1=(a-1)(a^{n-1}+a^{n-2}+\cdots+a+1)\geq (a-1)\cdot n\sqrt{a^{n-1}}=n\left(\sqrt{a^{n+1}}-\sqrt{a^{n-1}}\right)$$
onyqz
24.08.2024 10:06
same solution as @ehuseyinyigit
$$\frac{a^n-1}{n} = (a-1)\frac{\sum_{k=0}^{n-1}a^k}{n} \stackrel{\text{AM-GM}}{\geq} (a-1)\sqrt[n]{a^{\frac{n(n-1)}{2}}} = (a-1)a^{\frac{n-1}{2}} = \sqrt{a}^{n+1}-\sqrt{a}^{n-1} \, \square$$
simple motivation:
$a^n-1$ and the denominator $n$ invite us to factor out $a-1$, which would leave us with exactly $n$ terms. Hence we could abuse $\frac{1}{n}$ and $\text{AM-GM}$, which immediately gives the result.