Prove that for $x_i\in [0, 1],\,i=1,\dots,n$, the following inequality holds: $$n+x_1x_2...x_n \geq 1+x_1+x_2+...+x_n$$ I have tried Bernoulli's inequality which says $(1+x_1)(1+x_2)...(1+x_n)\geq 1+x_1+x_2+...+x_n$ for $x_i>-1$ and $x_i$ with the same sign.

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    $\begingroup$ At first i would solve the case $$n=2$$ $\endgroup$ – Dr. Sonnhard Graubner Jul 5 at 13:54
  • $\begingroup$ Welcome to Math Stack Exchange. Please use MathJax; currently it is not clear if xn means $x_n$ or $x^n$ $\endgroup$ – J. W. Tanner Jul 5 at 13:59

It's a linear inequality of $x_i$ for all $i$.

Since the linear function gets a minimal value for extreme value of the variable,

it's enough to check $x_i\in\{0,1\}$.

  • $\begingroup$ of $x_i$....... $\endgroup$ – mathworker21 Jul 5 at 14:12
  • $\begingroup$ @mathworker21 I fixed alredy. :) $\endgroup$ – Michael Rozenberg Jul 5 at 14:13
  • $\begingroup$ nope........... $\endgroup$ – mathworker21 Jul 5 at 14:14
  • $\begingroup$ @mathworker21 Thank you! :-D $\endgroup$ – Michael Rozenberg Jul 5 at 14:27

Put $x_i=1-a_i$ where $a_i\in[0,1]$, then $$n-x_1x_2\cdots x_n\ge1+x_1+x_2+\cdots+x_n$$ becomes equivalent to $$\sum a_i+\prod(1-a_i)\ge1\tag1 $$ But $$\prod(1-a_i)=1-\sum a_i+A\ge0$$ Thus $(1)$ becomes $$A\ge0$$ It is not difficult to prove that $A$ is non negative (it is a finite sequence $S_2-S_3+S_4-S_5+\cdots\pm S_n$ decreasing in absolute values of elementary symmetrics functions in $a_1,a_2,\cdots a_n$).


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