Question 86
Suppose positive real numbers $a, b, c, d$ satisfy $abcd = 1$. Prove
$$
\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} + \frac{9}{a+b+c+d} \ge \frac{25}{4}.
$$
$$
\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} + \frac{9}{a+b+c+d} \ge \frac{25}{4}.
$$