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

Question

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.

Answer 1

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: $$ \sum_k x_k^{S-x_k}>\sum_k \frac{x_k}{S}=1 $$

Answer 2

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 $$(ab+ac+bc)(a+b+c-1)+abc(3-a-b-c)>0.$$

Answer 3

Update

An ugly proof for $n=4$:

Due to symmetry, assume that $d = \max(a, b, c, d)$. Note that $v\mapsto u^v$ is decreasing on $v > 0$ provided $u\in (0,1)$, and that $x\mapsto x^y$ is convex on $x>0$ provided $y\ge 1$. We have \begin{align} a^{S-a} + b^{S-b} + c^{S-c} + d^{S-d} &\ge a^S + b^S + c^S + d^{S-d}\\ & \ge 3\Big(\frac{a+b+c}{3}\Big)^S + d^{a+b+c}. \end{align} Let $a+b+c = 3w$. Then $0 < w \le d < 1$. It suffices to prove that $$3w^{3w+d} + d^{3w} \ge 1.$$ It is easy to prove that $w^w > \mathrm{e}^{-1/\mathrm{e}}$ for $w\in (0, 1)$. Also, we have $d^{3w} \ge \frac{d}{d+ 3w - 3dw}$ by using $u^v \ge \frac{u}{u+v-uv}$ for $u>0, \ v\in (0, 1)$. Thus, it suffices to prove that $$3\mathrm{e}^{-3/\mathrm{e}}w^d + \frac{d}{d+ 3w - 3dw} \ge 1$$ or $$3\mathrm{e}^{-3/\mathrm{e}}w^{d-1} \ge \frac{3(1-d)}{d+ 3w - 3dw}.$$ For each fixed $d\in (0, 1)$, let $$f(w) = \ln \left(3\mathrm{e}^{-3/\mathrm{e}}w^{d-1}\right) - \ln \frac{3(1-d)}{d+ 3w - 3dw}, \quad w\in (0, d].$$ We have $$f'(w) = \frac{(1-d)d(3w-1)}{w(d+3w-3dw)}.$$ We split into two cases:

1) $d > \frac{1}{3}$: Note that $f'(w) < 0$ on $w\in (0, \frac{1}{3})$ and $f'(w)>0$ on $w\in (\frac{1}{3}, d]$. It suffices to prove that $f(\frac{1}{3}) \ge 0$ or $$g(d) = \ln \left(3\mathrm{e}^{-3/\mathrm{e}}\right) + (1-d)\ln 3 - \ln (1-d)\ge 0.$$ We have $g'(d) = -\ln 3 + \frac{1}{1-d} \ge -\ln 3 + \frac{1}{1 - 1/3} > 0$. It suffices to prove that $g(\frac{1}{3}) \ge 0$. It is true.

2) $d \le \frac{1}{3}$: Note that $f'(w) \le 0$ on $w\in (0, d]$. It suffices to prove that $f(d) \ge 0$ or $$\ln \left(3\mathrm{e}^{-3/\mathrm{e}}\right) - (1-d)\ln d - \ln \frac{3(1-d)}{4d - 3d^2}\ge 0$$ or $$\ln \left(3\mathrm{e}^{-3/\mathrm{e}}\right) + d\ln d - \ln \frac{3(1-d)}{4 - 3d}\ge 0.$$ It is easy to prove that $d\ln d \ge -\frac{1}{\mathrm{e}} + \frac{1}{2}\mathrm{e}\left(d-\frac{1}{\mathrm{e}}\right)^2$ and $ - \ln \frac{3(1-d)}{4 - 3d} \ge -\ln\tfrac{3}{4} + \frac{1}{4}d + \frac{7}{32}d^2$ for $d\in (0, \frac{1}{3}]$. Thus, it suffices to prove that $$\ln \left(3\mathrm{e}^{-3/\mathrm{e}}\right)-\frac{1}{\mathrm{e}} + \frac{1}{2}\mathrm{e}\left(d-\frac{1}{\mathrm{e}}\right)^2 -\ln\tfrac{3}{4} + \frac{1}{4}d + \frac{7}{32}d^2 \ge 0.$$ It is true.

We are done.