### 2008 IMO SL #A5

Let $a$, $b$, $c$, $d$ be positive real numbers such that $abcd = 1$ and $a + b + c + d > \dfrac{a}{b} + \dfrac{b}{c} + \dfrac{c}{d} + \dfrac{d}{a}$. Prove that

$a + b + c + d < \dfrac{b}{a} + \dfrac{c}{b} + \dfrac{d}{c} + \dfrac{a}{d}$

Solution 1:

By Cauchy-Schwarz, we have $$(a+b+c+d)(ab+bc+cd+da)>\left(\frac ab+\frac bc+\frac cd+\frac da\right)(ab+bc+cd+da)$$$$\ge(a+b+c+d)^2,$$which implies that $ab+bc+cd+da>a+b+c+d$. Again, by Cauchy-Schwarz, it follows that $$(a+b+c+d) \left(\frac ba+\frac cb+\frac dc+\frac ad\right) > \left(\frac ab+\frac bc+\frac cd+\frac da\right) \left(\frac dc+\frac ad+\frac ba+\frac cb\right)$$     $$\ge\left(\sqrt{\frac{da}{bc}}+\sqrt{\frac{ab}{cd}}+\sqrt{\frac{bc}{da}}+\sqrt{\frac{cd}{ab}}\right)^2$$$$=(da+ab+bc+cd)^2 \ge(a+b+c+d)^2,$$which implies $\frac ba+\frac cb+\frac dc+\frac ad>a+b+c+d$, as desired. $\square$

Solution 2:

Define $$S_1 = \frac ab +\frac bc +\frac cd +\frac da$$$$S_2 = \frac ba +\frac cb +\frac dc +\frac ad$$$$W =a+b+c+d$$. We claim that $3S_1+S_2 \geq 4W$ . Clearly, this finishes because then we have $S_2 \geq W + \underbrace{3(W-S_1)}_{>0} >W$.

To prove our claim, note that $$2 \cdot \frac ab + \frac bc + \frac ad \overset{\text{AM-GM}}{\geq} 4\sqrt [4]{\frac {a}{d}\cdot \frac {a^2}{b^2} \cdot \frac {b}{c}} \geq 4a$$. Cyclically summing up gives the desired claim, so we are done. $\square$

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