4
$\begingroup$

Let $x_{i}>0(i=1,2,\cdots,n),n\ge 3$. prove $$\sum_{i=1}^{n}\dfrac{1}{S-x_{i}}+\dfrac{n^n\displaystyle\prod_{i=1}^{n}x_{i}}{(n-1)(n-2)S^n}\ge\dfrac{n-1}{n-2},~~~~~~~~~S=\sum_{i=1}^{n}x_{i}$$

I try to prove Find a function like this$f$ such $$\sum_{i=1}^{n}\dfrac{1}{S-x_{i}}\ge f\left(\dfrac{S^n}{\displaystyle\prod_{i=1}^{n}x_{i}}\right)$$,then I want use $AM-GM$ inequality to prove it. But Until now, I couldn't find this function.

$\endgroup$
  • 10
    $\begingroup$ As stated like this, the inequality does not hold in general. If all $x_i$ are equal to, say, $a > 0$, then the inequality reduces to $a \leq 1$. Perhaps there is a missing $x_i$ in the numerator of the first sum? $\endgroup$ – P. Senden Apr 12 '18 at 16:04
1
+50
$\begingroup$

Expanding on the answer in the comments, a counterexample with $x_i > 0$ is given by $x_i = a$. Then $S = an$, and we have

$$\sum_{i=1}^{n}\dfrac{1}{an-a}+\dfrac{n^na^n}{(n-1)(n-2)a^nn^n}\ge\dfrac{n-1}{n-2}$$ $$\iff\dfrac{n}{a(n-1)}+\dfrac{1}{(n-1)(n-2)}\ge\dfrac{n-1}{n-2}$$ $$\iff\dfrac{n(n-2)}{a}+1\ge(n-1)^2$$ $$\iff n(n-2)+a\ge a(n-1)^2\\ \iff a \leq 1$$ since $n > 2$. So, picking any $a > 1$, say, $a = 2$ gives a counterexample.

$\endgroup$
0
$\begingroup$

P. Senden have written in the comments, that the issue inequality is wrong.

The counterexample is $$x= \left\{1, 2, 2\right\},\quad \dfrac{11}{12} < 2.$$

$\endgroup$
  • $\begingroup$ This isn't a counterexample, since the statement required $x_i > 0$. $\endgroup$ – B. Mehta Apr 20 '18 at 1:05
  • $\begingroup$ @B.Mehta Thanks! Fixed. $\endgroup$ – Yuri Negometyanov Apr 20 '18 at 15:18

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.