Show that ${\frac{x_1^{n}+x_2^{n}+...+x_n^{n}}{x_1x_2...x_n} + \frac{\sqrt[n]{x_1x_2...x_n}}{x_1+x_2+...+x_n}}\ge{n+\frac{1}{n}}$

Question

Let $n\in\mathbb{N}$ such that ${n}>{0}$.

Show that $${\frac{x_1^{n}+x_2^{n}+...+x_n^{n}}{x_1x_2...x_n}+\frac{\sqrt[n]{x_1x_2...x_n}}{x_1+x_2+...+x_n}}\ge{n+\frac{1}{n}}$$ when $x_k > 0, \forall k$.

I thought I can use AM GM inequality, but if I try to break it in two smaller pieces and apply AMGM I get a false affirmation for $\frac{\sqrt[n]{x_1x_2...x_n}}{\ x_1+\ x_2+...+x_n}$.

Any help please?

Answer 1

Since the inequality is homogeneous you can assume that $x_1x_2\cdots x_n=1.$ Let $x_1^n+x_2^n+\dots+x_n^n = A$ and $x_1+x_2+\dots+x_n = B$ for simplicity. Then, you have some easy inequalities: $$A\geq B\geq n$$ and $A\geq \dfrac{B^n}{n^{n-1}},$ the power mean inequality. Therefore, $$n\left(A+\dfrac{1}{B}\right)\geq \dfrac{B^n}{n^{n-2}}+\dfrac{n}{B}=n^2\left(\dfrac{B^n}{n^{n}}\right)+\dfrac{n}{B}\geq(n^2+1)\left(\dfrac{B^{n^3-1}}{n^{n^3-1}}\right)^{\frac{1}{n^2+1}}\geq n^2+1.$$

This proves your inequality and the equality is attained when all the variables are equal.

Answer 2

Supplement to dezdichado's proof:

WLOG, assume that $x_1 x_2 \cdots x_n = 1$. Let $B = x_1 + x_2 + \cdots + x_n$. We have $B \ge n$ by AM-GM.

Using AM-GM, we have $x_1^n + (n - 1) \ge n\sqrt[n]{x_1^n \cdot 1 \cdot 1 \cdots 1} = n x_1$ and hence $$x_1^n + x_2^n + \cdots + x_n^n \ge nB - n(n - 1).$$

It suffices to prove that $$n B - n(n - 1) + \frac{1}{B} \ge n + \frac{1}{n}$$ or $$nB - n(n - 1) - n \ge \frac{1}{n} - \frac{1}{B}$$ or $$n(B - n) \ge \frac{B - n}{nB}$$ which is true using $B \ge n$.

We are done.