Prove that:$\sum_{i=1}^n\frac{1}{x_i}-\sum_{i<j}\frac{1}{x_i+x_j}+\sum_{i<j<k}\frac{1}{x_i+x_j+x_k}-\cdots+(-1)^{n-1}\frac{1}{x_1+\ldots+x_n}>0.$?

Is there someone show me how do I prove this , I guess this inequality hold

only if $x=0$ .

Let $x_1,x_2,\ldots,x_n>0$. Prove that

$$\sum_{i=1}^n\frac{1}{x_i}-\sum_{i<j}\frac{1}{x_i+x_j}+\sum_{i<j<k}\frac{1}{x_i+x_j+x_k}-\cdots+(-1)^{n-1}\frac{1}{x_1+\ldots+x_n}>0.$$

• math.stackexchange.com/questions/1340809/… – math110 Jun 29 '15 at 15:26
• Note: stackexchang not post same question – math110 Jun 29 '15 at 15:45
• @Ani Fier, your guess is true f(x)\geq 0 can hold if x=0 – zeraoulia rafik Jun 29 '15 at 16:07
• @math110 I follow your link and see there is no answer there, so I added an answer there and didn't notice zeraoulia beat me for several minutes and give an answer here. – achille hui Jun 29 '15 at 16:12
• in:math.stackexchange.com/questions/1340809/… ,The inequality supposed true ,but in this question seek to prove it then there is a difference between 2 – zeraoulia rafik Jun 29 '15 at 16:13

for any $0 \le x \le 1$, we have $$1 - \prod_{i=1}^n (1 - x^{a_i}) \ge 0$$ and hence $$f(x) = \sum_{i=1}^n x^{a_i} - \sum_{i < j} x^{a_i + a_j} + \cdots + (-1)^{n-1} x^{a_1 + \cdots + a_n} \ge 0.$$ Now integrating $f(x)/x$ from $0$ to $1$, we get $$\int_0^1 \frac{f(x)}{x} dx = \sum_{i=1}^n \frac{1}{a_i} - \sum_{i<j} \frac{1}{a_i+a_j} + \cdots + (-1)^{n-1} \frac{1}{a_1 + \cdots + a_n} \ge 0.$$ Note that the equality in $f(x) \ge 0$ can hold only if $x = 0$. Thus the definite integral must always be positive