Is $\prod_{i=1}^{n}x_{i} > \prod_{i=1}^{n}y_{i} \implies \prod_{i=1}^{n}(x_{i}+k) > \prod_{i=1}^{n}(y_{i}+k)$ true?

Question

Is this true?

$$ x_{i} > 0, y_{i} > 0, k > 0, \\ x_{i} \in \mathbb R, y_{i} \in \mathbb R, k \in \mathbb R, \\ \prod_{i=1}^{n}x_{i} > \prod_{i=1}^{n}y_{i} \implies \\ \prod_{i=1}^{n}(x_{i}+k) > \prod_{i=1}^{n}(y_{i}+k) \\ $$

Answer 1

$\prod_{i=1}^{n}(x_{i}+k)$ is a polynomial in $k$ whose size for $k\to+\infty$ is $$k^n+k^{n-1}\sum_{i=1}^nx_i+O(k^{n-2}).$$

So we get a counterexample for large $k$ if $\sum_{i=1}^nx_i<\sum_{i=1}^ny_i$.

It's not too hard to find an example where $\sum_{i=1}^nx_i<\sum_{i=1}^ny_i$; here's a lazy way:

$x_1x_2=1\cdot1=2\cdot\frac12=y_1y_2$, and $1+1<2+\frac12$. Increase one of the $x_i$ a tiny bit so that $x_1x_2>y_1y_2$ but such that the inequality $\sum x_i<\sum y_i$ is preserved.