Prove by mathematical induction and determine when there is an equality

Question

Let's assume that numbers $x_1,x_2,...,x_n$ are of the same sign and that $x_1>-1, x_2>-1,...,x_n>-1$.Prove that $$(1+x_1)(1+x_2)...(1+x_n)\geq 1+x_1+x_2+...+x_n$$Determine when there is equality.
So I have few questions regarding this problem. I tried two inductions first one$$(1+x_1+x_2+...+x_n)(1+x_{n+1})\geq1+x_1+x_2+...+x_n+x_{n+1}$$ Which evaluates to$$x_1+x_2+...+x_n\geq0$$And we can't conclude anything from that right? How is that that if we just 'change sides' and write this $$(1+x_1)(1+x_2)...(1+x_n)(1+x_{n+1})\geq(1+x_1)(1+x_2)...(1+x_n)+x_{n+1}$$we get$$(1+x_1)(1+x_2)...(1+x_n)\geq1$$and that gives us nothing too? And when the equality holds? Can the answer be formal?

Answer 1

Hint:

This is Bernoulli's inequality. The trick for the inductive step is to multiply both sides of $\;(1+x_1)(1+x_2)...(1+x_n)\geq 1+x_1+x_2+...+x_n$ with $1+x_n$ as you did, expand the right-hand side and use that the $x_k$ all have the same sign.

Answer 2

I think your third inequality is wrong; it seems to me that if you distribute the product in $$(1+x_1+x_2+...+x_n)(1+x_{n+1})\geq1+x_1+x_2+...+x_n+x_{n+1},$$you should get that it is equivalent to $$(x_1+x_2+...+x_n)x_{n+1}\geq 0,$$ which holds because all the $x_i$ have the same sign.

Answer 3

We have $$(1+x_{ 1 })(1+x_{ 2 })...(1+x_{ n })\left( 1+x_{ n+1 } \right) \geq \left( 1+x_{ 1 }+x_{ 2 }+...+x_{ n } \right) \left( 1+x_{ n+1 } \right) =\\=1+x_{ 1 }+x_{ 2 }+...+x_{ n }+x_{ n+1 }+\left( x_{ 1 }+x_{ 2 }+...+x_{ n } \right) x_{ n+1 }\ge \\ \ge 1+x_{ 1 }+x_{ 2 }+...+x_{ n }+x_{ n+1 }$$

Answer 4

Consider the following: \begin{align} \prod_{i=1}^{k+1}(1+x_i)&= \prod_{i=1}^k(1+x_i)(1+x_{k+1})\tag{by definition}\\[1em] &\geq\left[1+\sum_{i=1}^kx_i\right](1+x_{k+1})\tag{by ind. hyp.}\\[1em] &= \sum_{i=1}^k x_i+x_{k+1}+1+\sum_{i=1}^kx_i\cdot x_{k+1}\tag{expand}\\[1em] &= \sum_{i=1}^{k+1}x_i+1+\sum_{i=1}^kx_i\cdot x_{k+1}\tag{rewrite}\\[1em] &\geq \sum_{i=1}^{k+1}x_i+1.\tag{all $x_i$ have same sign} \end{align} For equality, $x_i>-1$. Can $x_i=0$?