Proof of an inequality by induction: $(1 + x_1)(1 + x_2)...(1 + x_n) \ge 1 + x_1 + x_2 + ... + x_n$

Question

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?
...

Answer 1

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 $x_{n+1}(x_1+\ldots +x_n) \geq 0$.

Answer 2

Why not directly instead of cases?

$$\prod_{k=1}^n(1+x_k)=\prod_{k=1}^{n-1}(1+x_k)\cdot(1+x_n)\stackrel{\text{ind. hyp.}\,+\,(1+x_n)\ge0}\ge(1+x_1+\ldots+x_{n-1})(1+x_n)=$$

$$=1+\sum_{k=1}^n x_k+\sum_{k=1}^{n-1}\overbrace{x_kx_n}^{\ge 0}\ge1+\sum_{k=1}^n x_k$$

and we're done

Answer 3

There's no need for induction. Note that when you start increasing any $x_i$, the left side increases slower (or equally) because the product of the other factors besides $(x_i+1)$ is less or equal to $1$. So without loss of generality, all the variables are $0$ and the inequality holds.

Answer 4

Proof:

When $n=1$, it's obvious that $1+x_1 \geqslant 1+x_1$ holds.

Suppose $(1+x_1)(1+x_2)\ldots{}(1+x_k) \geqslant 1+x_1+x_2+\ldots{}+x_k$ holds when $n=k$.

\begin{align} (1+x_1)(1+x_2)\ldots{}(1+x_k)(1+x_{k+1}) &= (1+x_{k+1})\Big[(1+x_1)(1+x_2)\ldots{}(1+x_k)\Big] \\ &\geqslant (1+x_{k+1})(1+x_1+x_2+\ldots{}+x_k) \\ &=1 + x_{k+1} + (x_1+x_2+\ldots{}+x_k) + x_{k+1}(x_1+x_2+\ldots{}+x_k) \end{align}

Because $x_1,x_2,\ldots{},x_n \in [-1,0]$, $$\forall i, j \in \{1,2,\ldots{},n\} :\: x_i x_j \geqslant 0$$ $$x_{k+1}(x_1+x_2+\ldots{}+x_k) \geqslant 0$$

Furthermore, \begin{align} (1+x_1)(1+x_2)\ldots{}(1+x_k)(1+x_{k+1}) &\geqslant 1 + x_{k+1} + (x_1+x_2+\ldots{}+x_k) + x_{k+1}(x_1+x_2+\ldots{}+x_k) \\ &\geqslant 1 + x_1 + x_2 + \ldots{} + x_k + x_{k+1} \end{align}

Therefore, for all $n \in \mathbb{N}^{+}$, the proposition is always true.

Q.E.D.

Reference:

1. Bernoulli's inequality