How can one show that $|(a-b)(b-c)(c-d)(d-a)|\le\frac{abcd}{4}$ Given that a,b,c,d real numbers in the interval [1,2]
How can one show that :
$|(a-b)(b-c)(c-d)(d-a)|\le\frac{abcd}{4}$
And when equality occurs
I tried using AM GM but no results
 A: Let $a=1+\cos(2t_a)$, $b=1+\cos(2t_b)$, $c=1+\cos(2t_c)$ and $d=1+\cos(2t_d)$, where $t_a,t_b,t_c,t_d \in [0,\pi/4]$. This means we need to prove that
\begin{align}
\left \vert (\cos(2t_a)-\cos(2t_b)) (\cos(2t_b)-\cos(2t_c)) (\cos(2t_c)-\cos(2t_d)) (\cos(2t_d)-\cos(2t_a))\right \vert\\
\leq 4\cos^2(t_a)\cos^2(t_b)\cos^2(t_c)\cos^2(t_d)
\end{align}
Now we have that for all $t_i,t_j \in [0,\pi/4]$
$$\left \vert \cos(2t_i)-\cos(2t_j)\right \vert \leq \sqrt2 \cos(t_i)\cos(t_j)$$
which when multiplied out gives us what we want.

Equivalently, for $x,y \in [0,1]$, it suffices to show that
$$\vert x-y\vert \leq \sqrt{\dfrac{(1+x)(1+y)}2}$$
which is equivalent to proving that
$$2(x^2+y^2-2xy) \leq (1+x)(1+y) \iff 2x^2 + 2y^2 - 5xy - x-y - 1 \leq 0$$
which in turn is equivalent to proving that
$$(x-2y-1)(2x-y+1) \leq 0$$
which is true, since $2x-y+1 \geq 0$ and $x-2y-1 \leq 0$ for all $x,y \in [0,1]$. Applying the above inequality for pairs (i) $(a-1,b-1)$; (ii) $(b-1,c-1)$; (iii) $(c-1,d-1)$ and (iv) $(d-1,a-1)$; and multiplying them out, we obtain what we want.
A: $$\left|\prod_{cyc}(a-b)\right|=\sqrt{\prod_{cyc}(a-b)^2}\leq\sqrt{\prod_{cyc}\frac{ab}{2}}=\frac{abcd}{4}.$$
Done!
