Proof of a (simple?) inequality $x^n \geq \prod\limits_{i=1}^n (x+k_i)$

Question

I have a feeling that the following inequality should be very easy to prove:

$$ x^n \geq \prod_{i=1}^n{(x+k_i)},\quad\text{where } \sum_{i=1}^{n}{k_i}=0,\quad \text{and } x+k_i>0\text{ for all } i $$

(and the equality only holds when all the $k_i=0$).

It seems intuitively obvious (when $n=2$, a square has a greater area than a rectangle with the same perimeter, when $n=3$, a cube has greater volume than a rectangular prism with the same surface area, etc.) but I can't find an appropriately easy proof.

I think I can show it analytically by finding the local maximum for $f(x_1,\ldots,x_n)=\prod_{i=1}^n{x_i}$ within the box $\max{x_i}=r$ in the upper-right quadrant, but I feel like there should be a neat algebraic/geometric argument, since it's such an intuitive statement.

Answer 1

The AM-GM inequality gives us $$\prod_i (x+k_i)^{1/n} \leq {1\over n}\sum_i (x+k_i)=x.$$ Now take the $n$th power of both sides.

Answer 2

We can show it by induction: for $n=2$, we have for $a+b=0$: $x^2-(x+a)(x+b)=-(a+b)x-ab=-ab\geq 0$, since $ab\leq 0$. We assume that the result is true for $n$. Let $(k_1,\ldots,k_{n+1})$ such that $\sum_{j=1}^{n+1}k_j=0$. We can assume that $k_nk_{n+1}\leq 0$. We put $k_j'=k_j$ if $j\leq n-1$, and $k_n'=k_n+k_{n+1}$. We get by induction hypothesis $$x^n\geq \prod_{j=1}^n(x+k_j')=(x+k_n+k_{n+1})\prod_{j=1}^{n-1}(x+k_j),$$ so $$x^{n+1}\geq x(x+k_n+k_{n+1})\prod_{j=1}^{n-1}(x+k_j),$$ and we have to show that $x(x+k_n+k_{n+1})\geq (x+k_n)(x+k_{n+1})$, since $\prod_{j=1}^{n-1}(x+k_j)\geq 0$. But \begin{align*} x(x+k_n+k_{n+1})-(x+k_n)(x+k_{n+1})&=x(k_n+k_{n+1})-xk_{n+1}-xk_n-k_nk_{n+1}\\ &=-k_nk_{n+1}\geq 0. \end{align*}