Let $n \in \mathbb N^+$. Show that if $x_1, x_2, … , x_n$ are $n$ real numbers such that $-1 \le x_i \le 0$ for each $1 \le i \le n$, then $$(1 + x_1)(1 + x_2)…(1 + x_n) \ge 1 + x_1 + x_2 + … + x_n$$
I am using a proof by induction, but I am unable to complete it. I would appreciate any tips on how I could finish the proof.
Incomplete Proof
Let $P(n)$ be the proposition that $(1 + x_1)(1 + x_2)…(1 + x_n) \ge 1 + \sum_{i = 1} ^ n x_i$.
When $n = 1$, LHS = RHS = $1 + x_i$. Clearly $1 + x_i \ge 1 + x_i$ and so $P(1)$ is true.
Assume that $P(k)$ is true, i.e. $(1 + x_1)(1 + x_2)…(1 + x_k) \ge 1 + \sum_{i=1}^k x_i$.
Case 1
Suppose $x_{k + 1} = -1$.
Then $1 + x_{k + 1} = 0$, hence LHS = $(1 + x_1)(1 + x_2)…(1 + x_k)(1 + x_{k+1}) = 0$.
RHS = $1 + x_1 + x_2 + … + x_k – 1 = x_1 + x_2 + … + x_k$ which is always nonpositive since each $x_i$ is nonpositive.
Hence RHS $\le 0$ and so LHS $\ge$ RHS.
Case 2
Suppose $x_{k+1} = 0$.
Then $1 + x_{k+1} = 1$, hence multiplying it to LHS does not change anything.
Also, adding $x_{k+1}$ to RHS does not change anything.
Therefore, LHS $\ge$ RHS still holds.
Case 3
Suppose $-1 < x_{k+1} < 0$.
Then $0 < 1 + x_{k+1} < 1$.
Adding $x_{k+1}$ to RHS makes the RHS smaller since $x_{k+1}$ is negative.
If LHS $= 0$, then multiplying $1 + x_{k+1}$ to LHS does not change anything, and so LHS $\ge$ RHS holds.
If LHS > $0$, then multiplying $1 + x_{k+1}$ to LHS makes the LHS smaller since $0 < 1 + x_{k+1} < 1$.
In this case LHS $\ge$ RHS does not necessarily hold?
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Best Answer
We know that
$$(1+x_1)\cdots(1+x_n) \geq 1+x_1+\ldots +x_n$$
If we now multiply the inequality above by $(1+x_{n+1}) \geq 0$ we obtain
$$(1+x_1)\cdots(1+x_n)(1+x_{n+1}) \geq (1+x_1+\ldots +x_n)(1+x_{n+1})\\= (1+x_1+\ldots +x_n) + x_{n+1} + x_{n+1}(x_1+\ldots +x_n) \geq 1 + x_1 + \ldots + x_{n+1}$$
where the last inequality follows from the fact that $x_i \leq 0$ which implies $x_{n+1}(x_1+\ldots +x_n) \geq 0$.