Show that: $ x_{1}^n + x_{2}^n + · · · + x_{n}^n ≥ nx_{1} x_{2} · · · x_{n} $.

Question

Let $ x_{1}, x_{1}, x_{2}, · · ·, x_{n} $ be positive reals. Show that: $ x_{1}^n + x_{2}^n + · · · + x_{n}^n ≥ nx_{1} x_{2} · · · x_{n} $. If $ x_{1} x_{2} · · · x_{n} = 1 $ then $ x_{1} + x_{2} + · · · + x_{n} ≥ n $.

I am trying to do this problem using Holder's inequality. Also I want to use a reasoning similar to the following exercise: $ x_1, x_2, ..., x_n \in \mathbb R $ are nonegative and $ k \in \mathbb R $, $ k \geq 1 $. Prove that if $ x_1 + x_2 + ... + x_n = n $, then $ x_1^k + x_2^k + ... + x_n^k \geq n $. I tried to find the smallest value of $ x_1^k + ... + x_n^k $ and it turned out to be $ n $ for $ x_1 = ... = x_n = 1 $. Am I doing this correct?

Answer 1

Use AM-GM:$$ \frac{x_1^n+x_2^n+x_3^n+...+x_n^n}{n} \ge (x_1^n x_2^n x_3^n....x_n^n)^{1/n}$$

Answer 2

The first it's AM-GM, which is just Jensen for a concave function $\ln$: $$\ln\frac{x_1+...+x_n}{n}\geq \frac{\ln{x_1}+...+\ln{x_n}}{n},$$ which gives $$\ln\frac{x_1+...+x_n}{n}\geq \ln\sqrt[n]{x_1\cdot...\cdot x_n}$$ or $$\frac{x_1+...+x_n}{n}\geq \sqrt[n]{x_1\cdot...\cdot x_n},$$ which for $x_1...x_n=1$ gives: $$x_1+...+x_n\geq n.$$ The second it's Power Mean inequality, which is Jensen again for the convex function $f(x)=x^k$: $$\frac{x_1^k+...+x_n^k}{n}\geq\left(\frac{x_1+...+x_n}{n}\right)^k.$$