# Prove that for x in [0,1] the inequality..

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.

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

It's a linear inequality of $$x_i$$ for all $$i$$.
it's enough to check $$x_i\in\{0,1\}$$.
• of $x_i$....... – mathworker21 Jul 5 at 14:12
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 $$a_1,a_2,\cdots a_n$$).