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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 $$ \frac{a+b}{b+c} + \frac{c+d}{d+a} ≤ 4\Big(\frac{a+c}{b+d}\Big) $$

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Please use $LaTeX$. –  ᴊ ᴀ s ᴏ ɴ Oct 18 '12 at 4:32

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up vote 3 down vote accepted

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 $$ \frac {a + b} {b + c} + \frac{c + d} {d + a} \leq a + c \leq \frac 4 {b + d} (a + c) $$

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