<s>[INST] How to prove $\prod\limits_{i=1}^{n}(x-4i+2)(x-4i+1)>\prod\limits_{i=1}^{n}(x-4i+3)(x-4i)$ for all $x\in\mathbb{R}$? I would like to prove that for $n\in\mathbb{N}$ we have $f_n(x):=\prod\limits_{r=1}^{n}(x-4r+2)(x-4r+1)>\prod\limits_{r=1}^{n}(x-4r+3)(x-4r)=:g_n(x)$ for all $x\in\mathbb{R}$ (actually it would suffice for $n$ even). My attempt was to form pairs on each side in such a way that we always obtain the same function plus a constant, i.e. for $n=2k$ even:
$$
f_n(x)=(x-2)(x-8k+1)\cdot(x-3)(x-4n+2)\cdot...\cdot(x-4k-2)(x-4k+1)=\\((x-4k-0.5)^2-(4k-1.5)^2)\cdot((x-4k-0.5)^2-(4k-2.5)^2)\cdot...\cdot((x-4k-0.5)^2-(1.5)^2)
$$
And similarly for $g_n(x)$. If we then substitute $x=0.5y+4k+0.5$ we see that the inequality is equivalent to:
$$
(y^2-3^2)(y^2-5^2)\cdot...\cdot(y^2-(2k-5)^2)(y^2-(2k-3)^2)>(y^2-1^2)(y^2-7^2)\cdot...\cdot(y^2-(2k-7)^2)(y^2-(2k-1)^2)
$$
for all $y\in\mathbb{R}$. I think now it would be enough to expand both sides and probably we would end up with a polynomial in $y^2$ where all coefficients are positive. However, computing all the coefficients will be quite tedious. How to prove it more elegantly?
Edit:
As several comments and answers use this same idea: simply reducing it to $(x-4r+2)(x-4r+1)=(x-4r)^2+3(x-4r)+2>(x-4r)^2+3(x-4r)=(x-4r+3)(x-4r)$ doesn't work as $a>c,b>d$ doesn't imply $ab>cd$ in general. But it was my first reflex too :)
 [/INST] $\begin{array}\\
(x-4i+2)(x-4i+1)-(x-4i+3)(x-4i)
&=((x-4i)^2+3(x-4i)+2)-((x-4i)^2+3(x-4i))\\
&=2\\
\end{array}
$
so each individual term is greater,
so their product is greater.
</s>