19-th Macedonian Mathematical Olympiad

By multiplying $1+bc+cd+da$ and $1+ab$ together we get $$\begin{gathered} (1+bc+cd+da)(1+ab)=1+bc+cd+da+ab+a{b}^{2}c+abcd+a^{2}bd= \\ =2+ab+bc+cd+da+\frac{b}{d}+\frac{a}{c} \end{gathered}$$ From the inequality between the arithmetic and the geometric mean for the positive numbers $\frac{b}{d}$ and $\frac{a}{c}$, and from the equality $abcd=1$ we get $\frac{b}{d}+\frac{a}{c}\ge 2\sqrt{\frac{ab}{cd}}=2ab$, and hence it holds that $(1+bc+cd+da)(1+ab)\ge 2+2ab+ab+bc+cd+da$ i.e. $$1+bc+cd+da\ge 2+\frac{ab+bc+cd+da}{1+ab}$$ or $$\frac{1+ab}{ab+bc+cd+da}\ge \frac{1}{bc+cd+da-1}(1)$$ Analogously we get $$\frac{1+bc}{ab+bc+cd+da}\ge \frac{1}{ab+cd+da-1}(2)$$ $$\frac{1+da}{ab+bc+cd+da}\ge \frac{1}{ab+bc+cd-1}(3)$$ and $$\frac{1+da}{ab+bc+cd+da}\ge \frac{1}{ab+bc+cd-1}(4)$$ By adding (1), (2), (3) and (4) together we get the inequality $$\frac{4+ab+bc+cd+da}{ab+bc+cd+da}\ge \frac{1}{bc+cd+da-1}+\frac{1}{ab+cd+da-1}+\frac{1}{ab+bc+da-1}+\frac{1}{ab+bc+cd-1}$$ Since $ab+bc+cd+da\ge 4\sqrt[4]{(abcd{)}^2}=4$ it follows that $$\frac{1}{bc+cd+da-1}+\frac{1}{ab+cd+da-1}+\frac{1}{ab+bc+da-1}+\frac{1}{ab+bc+cd-1}\le 1+\frac{4}{ab+bc+cd+da}\le 2$$ which was to be proven.

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Alternative solution.

We will introduce the substitutions $ab=p$ and $bc=q$. According to that, $cd=\frac{1}{p}$ and $ad=\frac{1}{q}$, so the given inequality is equivalent to the inequality $$\frac{1}{q+\frac{1}{p}+\frac{1}{q}-1}+\frac{1}{p+\frac{1}{p}+\frac{1}{q}-1}+\frac{1}{p+q+\frac{1}{q}-1}+\frac{1}{p+q+\frac{1}{p}-1}\le 2$$ On the other hand, if we use the inequalities $p+\frac{1}{p}\ge 2$ and $q+\frac{1}{q}\ge 2$ we get $$\begin{array}{rl}& \frac{1}{q+\frac{1}{p}+\frac{1}{q}-1}+\frac{1}{p+\frac{1}{p}+\frac{1}{q}-1}+\frac{1}{p+q+\frac{1}{q}-1}+\frac{1}{p+q+\frac{1}{p}-1}\\ & \le \frac{1}{1+\frac{1}{p}}+\frac{1}{1+\frac{1}{q}}+\frac{1}{1+p}+\frac{1}{1+q}=\\ & =\frac{p}{p+1}+\frac{1}{p+1}+\frac{q}{q+1}+\frac{1}{q+1}=2.\end{array}$$