Rickyminer wrote:
By Surányi's Inequality and Cauchy-Schwarz,
\begin{align*} (n-1) \sum_{i=1}^n x_i + n \left( \prod_{i=1}^n x_i \right)^{\frac{1}{n}} & \ge \left( \sum_{i=1}^n x_i^{\frac{1}{n}} \right) \left( \sum_{i=1}^n x_i^{\frac{n-1}{n}} \right) \\ & \ge \left( \sum_{i=1}^n x_i^{\frac{1}{2}} \right)^2. \end{align*}
We don't need to use Suranyi's Inequality here. We will prove that for any $n$ positive reals $a_1, a_2, \cdots, a_n$, the following inequality holds.
$$(n-1) \sum_{i=1}^{n} a_i^2 + n \sqrt[n]{\prod_{i=1}^{n} a_i^2} \ge \left( \sum_{i=1}^{n} a_i \right)^2$$Let $x_i = \log a_i$ for all $1 \le i \le n$, and $f = \exp$. Let $\bar{x} = \dfrac{x_1 + x_2 + \cdots + x_n}{n}$. The inequality is equivalent with the following.
$$(n-1)(f(2x_1) + f(2x_2) + \cdots + f(2x_n)) + nf(2\bar{x}) \ge f(2x_1) + f(2x_2) + \cdots + f(2x_n) + 2\sum_{1 \le i < j \le n}f(x_i+x_j)$$We use Karamata here, as $f$ is concave up. Without loss of generality, assume that $x_1 \ge x_2 \ge \cdots \ge x_n$. It suffices to prove that
$$(\underbrace{2x_1, \cdots, 2x_1}_{n-2 \text{ times}}, \underbrace{2x_2, \cdots, 2x_2}_{n-2 \text{ times}}, \cdots, \underbrace{2x_n, \cdots, 2x_n}_{n-2 \text{ times}}, \underbrace{2\bar{x}, \cdots, 2\bar{x}}_{n \text{ times}}) \text{ majorizes } (x_1 + x_2, x_1 + x_2, \cdots, x_{n-1} + x_n, x_{n-1} + x_n)$$Since a sequence $\textbf{a}$ majorizes a sequence $\textbf{b}$ if and only if $\sum_{i=1}^{n} (|a_i-x| - |b_i-x| ) \ge 0$ for any real $x$, we only need to prove that for any real $r$,
$$2(n-2) \sum_{i=1}^{n} | x_i - r| + 2n | \bar{x} - r| \ge 4 \sum_{1 \le i < j \le n}|x_i+x_j - 2r|$$By shifting the sequence, we only need to consider $r = 0$, which is equivalent to a well known inequality,
$$(n-2)\sum_{i=1}^{n}|x_i| + |\sum_{i=1}^{n} x_i| \ge 2 \sum_{1 \le i < j \le n} |x_i + x_j|$$