3
$\begingroup$

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

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

$\endgroup$
  • $\begingroup$ math.stackexchange.com/questions/1340809/… $\endgroup$ – math110 Jun 29 '15 at 15:26
  • $\begingroup$ Note: stackexchang not post same question $\endgroup$ – math110 Jun 29 '15 at 15:45
  • $\begingroup$ @Ani Fier, your guess is true f(x)\geq 0 can hold if x=0 $\endgroup$ – zeraoulia rafik Jun 29 '15 at 16:07
  • $\begingroup$ @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. $\endgroup$ – achille hui Jun 29 '15 at 16:12
  • $\begingroup$ 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 $\endgroup$ – zeraoulia rafik Jun 29 '15 at 16:13
11
$\begingroup$

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

$\endgroup$
  • $\begingroup$ It's nice solution!+1 $\endgroup$ – math110 Jun 29 '15 at 16:06
  • $\begingroup$ @zeraoulia rafik , thank you for this answer . $\endgroup$ – Ani Fier Jun 29 '15 at 16:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.