Question 73
If $a$, $b$, $c$, $d$ are positive real numbers such that $abcd = 1$ then prove that the inequality
$$
\frac{1}{bc+cd+da-1} + \frac{1}{ab+cd+da-1} + \frac{1}{ab+bc+da-1} + \frac{1}{ab+bc+cd-1} \le 2
$$
holds.
$$
\frac{1}{bc+cd+da-1} + \frac{1}{ab+cd+da-1} + \frac{1}{ab+bc+da-1} + \frac{1}{ab+bc+cd-1} \le 2
$$
holds.