How to prove this $\sum_{i=1}^{n}(x_{i})^{S-x_{i}}>1?$ Question:

Let $x_{i} \in (0,1),i=1,2,\cdots,n$. Show that
$$
  x_{1}^{S-x_{1}}+x_{2}^{S-x_{2}}+\cdots+x_{n}^{S-x_{n}}>1
$$
where $S=x_{1}+x_{2}+\cdots+x_{n}$.

I have proved when $n=2$,because it use this Bernoulli's inequality
$$
  (1+x)^a\le 1+ax,0<a\le 1,x>-1
$$
so we have
$$
  x^y=\dfrac{1}{(1/x)^y}=\dfrac{1}{\left(1+\frac{1-x}{x}\right)^y}\ge\dfrac{1}{1+\frac{1-x}{x}\cdot y}=\dfrac{x}{x+y-xy}>\dfrac{x}{x+y}
$$
and simaler we have
$$
  y^x>\dfrac{y}{x+y}
$$
so we have
$$
  x^y+y^x>\dfrac{x}{x+y}+\dfrac{y}{x+y}=1
$$

Edit: Now the Mr Michael Rozenberg has prove when $n=3$ and MR Czylabson Asa has prove $n\ge 6$ this inequality can't hold, so how to prove $n=4,5?$
Thanks.


Let $n>2$ and $x_k=\frac{2}{n},\ k=1,\ldots,n$. Then $S=2$, and the inequality takes the form:
$$
n\left( \frac{2}{n}\right)^{2-\frac{2}{n}}>1 \ \ \Leftrightarrow\\
2-\frac{2}{n}< \frac{\log\left(\frac{1}{n}\right)}{\log\left(\frac{2}{n}\right)}=\frac{\log(n)}{\log(n)-\log(2)},
$$
which is not true for $n\ge 6\ $. So, try to focus for the $n=3,4,5$ part, I was unable to find counterexamples with computer in those cases...

update:

There is an easy way to prove the inequality for all $n\ge 2$ with the assumption, that $S=\sum_k x_k\le 1$, with the help of this elementary Lemma: 
$$
u^v>\frac{u}{u+v}\ \ \text{if }\ \ 0<u\text{ and } 0<v<1
$$
Here it is:
$$
\sum_k x_k^{S-x_k}>\sum_k \frac{x_k}{S}=1
$$
