Let $a_1>a_2>...>a_r$ be positive real numbers. Compute $\lim_{n\to\infty}(a_1^n+a_2^n+...+a_r^n)^{1/n}$.
My approach: Since $a_i$ are positive real numbers, this implies that by AM-GM inequality we have $$\frac{a_1^n+a_2^n+...+a_r^n}{r}\ge(a_1.a_2.\cdots.a_r)^{n/r}$$ $$\implies a_1^n+a_2^n+...+a_r^n\ge r(a_1.a_2.\cdots.a_r)^{n/r}$$ $$\implies (a_1^n+a_2^n+...+a_r^n)^{1/n}\ge r^{1/n}(a_1.a_2.\cdots.a_r)^{1/r}...(1)$$
Again since $a_i$ are positive real numbers, this implies that by RMS-AM inequality we have $$\left(\frac{a_1^{2n}+a_2^{2n}+...+a_r^{2n}}{r}\right)^{1/2}\ge \frac{a_1^n+a_2^n+...+a_r^n}{r}$$ $$r^{1/2}\left(a_1^{2n}+a_2^{2n}+...+a_r^{2n}\right)^{1/2}\ge a_1^{n}+a_2^{n}+...+a_r^{n}$$ $$r^{1/2n}\left(a_1^{2n}+a_2^{2n}+...+a_r^{2n}\right)^{1/2n}\ge \left(a_1^{n}+a_2^{n}+...+a_r^{n}\right)^{1/n}...(2)$$
Now since $a_1\ge a_i,$ $\forall r,$ implies that $a_1^{2n}\ge a_i^{2n}, \forall r,$ which in turn helps us in concluding that $$\sum_{i=1}^ra_1^{2n}\ge \sum_{i=1}^ra_i^{2n}$$ $$\implies r.a_1^{2n}\ge a_1^{2n}+a_2^{2n}+...+a_r^{2n}$$ $$\implies r^{1/2n}.a_1\ge (a_1^{2n}+a_2^{2n}+...+a_r^{2n})^{1/2n}$$ $$\implies r^{1/n}.a_1\ge r^{1/2n}(a_1^{2n}+a_2^{2n}+...+a_r^{2n})^{1/2n}...(3)$$
Combining $(1),(2)$ and $(3)$, we can conclude that $$r^{1/n}.a_1\ge r^{1/2n}(a_1^{2n}+a_2^{2n}+...+a_r^{2n})^{1/2n}\ge \left(a_1^{n}+a_2^{n}+...+a_r^{n}\right)^{1/n}\ge r^{1/n}(a_1.a_2.\cdots.a_r)^{1/r}$$
$$\implies r^{1/n}.a_1\ge \left(a_1^{n}+a_2^{n}+...+a_r^{n}\right)^{1/n}\ge r^{1/n}(a_1.a_2.\cdots.a_r)^{1/r}$$
$$\implies \lim_{n\to\infty}r^{1/n}.a_1\ge \lim_{n\to\infty}\left(a_1^{n}+a_2^{n}+...+a_r^{n}\right)^{1/n}\ge \lim_{n\to\infty}r^{1/n}(a_1.a_2.\cdots.a_r)^{1/r}$$
$$\implies a_1\ge \lim_{n\to\infty}\left(a_1^{n}+a_2^{n}+...+a_r^{n}\right)^{1/n}\ge (a_1.a_2.\cdots.a_r)^{1/r}$$
How to proceed after this?
