Find this least postive real $a$ let $n$ be an integer with $n\ge 2$,Find the least postive real number $a$ such that
$$(n-1)\sqrt{1+a\left(n\sum_{i=1}^{n}x^2_{i}-\left(\sum_{i=1}^{n}x_{i}\right)^2\right)}+\prod_{i=1}^{n}x_{i}\ge\sum_{i=1}^{n}x_{i}$$
I try let $$x_{1}=x_{2}=\cdots=x_{n-1}=x,x_{n}=0$$
then
$$(n-1)\sqrt{1+a(n-1)x^2}\ge (n-1)x$$
$$\sqrt{1+a(n-1)x^2}\ge x$$
so
$$\dfrac{1}{x^2}+a(n-1)\ge 1$$
let $x\to+\infty$,we have
$$a\ge\dfrac{1}{n-1}$$
so I think the minumum of the value $a\ge\dfrac{1}{n-1}$.
then I can't prove $$(n-1)\sqrt{1+\dfrac{1}{n-1}\left(n\sum_{i=1}^{n}x^2_{i}-\left(\sum_{i=1}^{n}x_{i}\right)^2\right)}+\prod_{i=1}^{n}x_{i}\ge\sum_{i=1}^{n}x_{i}$$ .Thanks
 A: Problem: Let $n\ge 3$. Let $x_i \ge 0, \forall i$. Prove that
$$(n-1)\sqrt{1+\dfrac{1}{n-1}\left(n\sum_{i=1}^{n}x^2_{i}-\left(\sum_{i=1}^{n}x_{i}\right)^2\right)}
+\prod_{i=1}^{n}x_{i}\ge\sum_{i=1}^{n}x_{i}.$$
Proof:
By Vasc's Equal Variable Theorem [1, Corollary 1.8], we only need to prove the case when
either $x_1 = 0$ or $0 < x_1 \le x_2 = x_3 = \cdots = x_n$.

*

*$x_1 = 0$:

It suffices to prove that
$$(n-1)\sqrt{1+\dfrac{1}{n-1}\left(n\sum_{i=2}^{n}x^2_{i}-\left(\sum_{i=2}^{n}x_{i}\right)^2\right)}
\ge\sum_{i=2}^{n}x_{i}$$
that is
$$\frac{(n - 1)^2}{n} + (n - 1)\sum_{i=2}^{n}x^2_{i} \ge \left(\sum_{i=2}^{n}x_{i}\right)^2$$
which is true since $(n - 1)\sum_{i=2}^{n}x^2_{i} \ge \left(\sum_{i=2}^{n}x_{i}\right)^2$ by AM-QM.


*$0 < x_1 \le x_2 = x_3 = \cdots = x_n$:

Let $x_1 = a, x_2 = b$. It suffices to prove that, for all $0 < a \le b$,
$$(n - 1)\sqrt{1 + (b - a)^2} + a b^{n - 1} \ge a + (n - 1) b$$
that is
$$\frac{\sqrt{1 + (b - a)^2} - b}{a} \ge \frac{1 - b^{n - 1}}{n - 1}.$$
It is easy to prove that $x \mapsto \frac{1 - b^x}{x}$ is non-increasing on $[1, \infty)$.
Thus, it suffices to prove the case when $n = 2$, that is
$$\frac{\sqrt{1 + (b - a)^2} - b}{a} \ge 1 - b$$
which is true (easy to prove).
We are done.
Reference
[1] Vasile Cirtoaje, “The Equal Variable Method”, J. Inequal. Pure and Appl. Math., 8(1), 2007.
https://www.emis.de/journals/JIPAM/images/059_06_JIPAM/059_06.pdf
