$\frac{a+b}{b+c} + \frac{c+d}{d+a} ≤ 4(\frac{a+c}{b+d}) ; a , b , c , d ∈ [1 , 2]$ Is it true that
if all the real numbers $a , b , c , d$ are from the closed interval $[1 , 2]$ then we always have the inequality 
$$
\frac{a+b}{b+c} + \frac{c+d}{d+a} ≤ 4\Big(\frac{a+c}{b+d}\Big)
$$

Since $a, c \geq 1$, we have
$$
a - \frac {a + b} {b + c} = \frac{ab + ac - a - b} {b + c} = \frac {b(a - 1) + a(c - 1)} {b + c} \geq 0
$$
A similar argument leads to
$$
c - \frac{c + d} {d + a} \geq 0
$$
Finally $b + d\leq 4$ implies
$$
\frac {a + b} {b + c} + \frac{c + d} {d + a} \leq a + c \leq \frac 4 {b + d} (a + c)
$$
