Balkan Mathematical Olympiad

First we observe that $$16ac+\frac{a}{c^{2}b}+\frac{16c}{a^{2}d}+\frac{4}{ac}=\left(a+\frac{1}{a^{2}d}\right)\left(16c+\frac{1}{c^{2}b}\right)$$ $$\text{and }bd+\frac{b}{256d^{2}c}+\frac{d}{b^{2}a}+\frac{1}{64bd}=\left(b+\frac{1}{b^{2}a}\right)\left(d+\frac{1}{256d^{2}c}\right).$$ Thus we have equivalent inequality $(a+\frac{1}{a^{2}d})(b+\frac{1}{b^{2}a})(16c+\frac{1}{c^{2}b})(d+\frac{1}{256d^{2}c})\ge \frac{81}{4}$, which will be proved. From AM-GM for the positive numbers $\frac{a}{2}$, $\frac{a}{2}$, $\frac{1}{a^{2}d}$ we get $a+\frac{1}{a^{2}d}\ge 3\sqrt[3]{\frac{1}{4d}}$. In the same way $b+\frac{1}{b^{2}a}\ge 3\sqrt[3]{\frac{1}{4a}}$.

From AM-GM for the positive numbers $8c$, $8c$, $\frac{1}{c^{2}b}$ we get $16c+\frac{1}{c^{2}b}\ge 12\sqrt[3]{\frac{1}{b}}$. $$\text{Also }d+\frac{1}{256d^{2}c}\ge \frac{3}{8}\sqrt[3]{\frac{1}{2c}}.$$ $$\text{Finally we have }(a+\frac{1}{a^{2}d})(b+\frac{1}{b^{2}a})(16c+\frac{1}{c^{2}b})(d+\frac{1}{256d^{2}c})\ge \frac{81}{4}.$$ $$\text{The equality holds when }a=2,b=1,c=\frac{1}{2}\text{ and }d=\frac{1}{4}.$$