<s>[INST] 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.

 [/INST] For $n=3$ we can use your work and the Canhang's idea.

Let $\{a,b,c\}\subset(0,1).$ Prove that:
  $$a^{b+c}+b^{a+c}+c^{a+b}>1.$$

Proof.
Let $a+b+c\leq1.$ 
Thus, by Bernoulli 
$$\sum_{cyc}a^{b+c}=\sum_{cyc}\frac{1}{\left(1+\frac{1}{a}-1\right)^{b+c}}\geq\sum_{cyc}\frac{1}{1+\left(\frac{1}{a}-1\right)(b+c)}>\sum_{cyc}\frac{1}{1+\frac{b+c}{a}}=1.$$
Let $a+b+c\geq1.$
Thus, by Bernoulli again and by C-S we obtain:
$$\sum_{cyc}a^{b+c}=\sum_{cyc}\frac{1}{\left(1+\frac{1}{a}-1\right)^c\left(1+\frac{1}{a}-1\right)^b}\geq\sum_{cyc}\frac{1}{\left(1+\left(\frac{1}{a}-1\right)c\right)\left(1+\left(\frac{1}{a}-1\right)b\right)}=$$
$$=\sum_{cyc}\frac{a^2}{(a+b-ab)(a+c-ac)}\geq\frac{(a+b+c)^2}{\sum\limits_{cyc}(a+b-ab)(a+c-ac)}>1$$ because the last inequality it's just
$$(ab+ac+bc)(a+b+c-1)+abc(3-a-b-c)>0.$$
</s>