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 :)
 A: Let $a_r(x)=(x-(4r-2))(x-(4r-3)), b_r(x)=(x-(4r-1))(x-4r),
c_r(x)=\frac{a_r(x)}{b_r(x)}$,
and $A_n(x)=\prod_{r=1}^{n}a_r(x),B_n(x)=\prod_{r=1}^{n}b_r(x)$.
We wish to show that $A_n(x)>B_n(x)$ for any $n$ and $x$. 
Note that when $n=1$, the result follows from $A_1-B_1=2$. Suppose now that
$n\geq 2$.  For $k\in[|1,n|]$, let $I_k=[4k-2,4k-1], J_k=[4k-3,4k]$.
If $x$ is outside all the $J_k(1\leq k \leq n)$, then all the $a_r(x),b_r(x),c_r(x)$ are positive. From the $n=1$ case, we have $a_r>b_r$, whence $c_r>1$
and $\frac{A_n}{B_n}=\prod_{r=1}^n c_r >1$, so we are done.
If $x$ is in some $J_k\setminus I_k(1\leq k \leq n)$, then $A_n\geq 0 \geq B_n$
and $A_n$ and $B_n$ are not both zero, so we are done. 
The only case left is therefore when $x$ is in some $I_k(1\leq k \leq n)$. And on that interval, $A_n$ and $B_n$ are both negative ; our job is then to show that 
$\frac{A_n}{B_n}=\prod_{r=1}^n c_r$ is  $\leq 1$ on $I_k$.
It is easy to see that the derivative of $c_r$ vanishes at $4r-\frac{3}{2}$ and
nowhere else. It follows that $c_r$ attains its maximum value on $I_k$ at a point
$m_k$ defined by
$$
m_k=\left\lbrace\begin{array}{lcl}
4k-2 & \text{when} & r \lt k, \\
4k-\frac{3}{2} & \text{when} & r=k,\\
4k-1 & \text{when} & r \gt k. \\
\end{array}\right.
$$
So
$$
\begin{array}{lcl}
\prod_{r=1}^n c_r 
& = & \bigg(\prod_{r=1}^{k-1} c_r \bigg) c_k \bigg( \prod_{r=k+1}^{n} c_r \bigg) \\
&\leq & \bigg(\prod_{r=1}^{k-1} c_r(4k-2) \bigg) c_k(4k-\frac{3}{2}) \bigg(  \prod_{r=k+1}^{n} c_r(4k-1) \bigg) \\
&\leq & \bigg(\prod_{t=1}^{k-1} c_{k-t}(4k-2) \bigg) c_k(4k-\frac{3}{2}) \bigg( \prod_{t=1}^{n} c_{k+t}(4k-1) \bigg) \\
& = & \bigg(\prod_{t=1}^{k-1}  \frac{8t^2-2t}{8t^2-2t-1}\bigg) \frac{1}{9} \bigg( \prod_{t=1}^{n} \frac{8t^2-2t}{8t^2-2t-1} \bigg) \\
& \leq  &  \frac{1}{9} \bigg( \prod_{t=1}^{\infty} \frac{8t^2-2t}{8t^2-2t-1} \bigg)^2 \
( \ \text{because} \ \frac{8t^2-2t}{8t^2-2t-1} >1 ) \\
& \leq  &  \frac{1}{9} \bigg( \prod_{t=1}^{\infty} \frac{t^2+2t+1}{t^2+2t} \bigg)^2 \
( \ \text{because} \ 8t^2-2t-1\geq t^2+2t) \\
& =  &  \frac{1}{9} \bigg( \prod_{t=1}^{\infty} \frac{t+1}{t}\frac{t+1}{t+2} \bigg)^2 \\
& \leq  &  \frac{4}{9} \ ( \ \text{telescoping product} \ )
 \\
 \\
\end{array}
$$
This concludes the proof. 
A: $\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.
