Estonian Math Competitions

Multiplying all equations gives $$\left(a^2+\frac{1}{b^2}\right)\left(b^2+\frac{4}{c^2}\right)\left(c^2+\frac{16}{d^2}\right)\left(d^2+\frac{4}{a^2}\right)=2^8.$$ By AM-GM, $a^2+\frac{1}{b^2}\ge 2\cdot \frac{a}{b}$, where the equality holds if and only if $a=\frac{1}{b}$. Similarly $b^2+\frac{4}{c^2}\ge 4\cdot \frac{b}{c}$ (equality if and only if $b=\frac{2}{c}$), $c^2+\frac{16}{d^2}\ge 8\cdot \frac{c}{d}$ (equality if and only if $c=\frac{4}{d}$) and $d^2+\frac{4}{a^2}\ge 4\cdot \frac{d}{a}$ (equality if and only if $d=\frac{2}{a}$). Multiplying the obtained four inequalities gives $$\left(a^2+\frac{1}{b^2}\right)\left(b^2+\frac{4}{c^2}\right)\left(c^2+\frac{16}{d^2}\right)\left(d^2+\frac{4}{a^2}\right)\ge 2\cdot \frac{a}{b}\cdot 4\cdot \frac{b}{c}\cdot 8\cdot \frac{c}{d}\cdot 4\cdot \frac{d}{a}=2^8.$$ The resulting inequality must hold as equality by the first step of the solution. This is possible only if all four inequalities hold as equalities, whence $$\{\begin{array}{l}a=\frac{1}{b},\\ b=\frac{2}{c},\\ c=\frac{4}{d},\\ d=\frac{2}{a}.\end{array}$$ By multiplying the equations of this system, we get $abcd=\frac{16}{abcd}$, whence $(abcd{)}^2=16$. As $a,b,c$ and $d$ are positive, the only possibility is $abcd=4$.

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Alternative solution.

By introducing $a=\frac{x}{2}$, $b=2y$, $c=z$, $d=4t$, rewrite the system as $$\{\begin{array}{l}\frac{x^2}{4}+\frac{1}{4y^2}=\frac{1}{2},\\ 4y^2+\frac{4}{z^2}=8,\\ z^2+\frac{16}{16t^2}=2,\\ 16t^2+\frac{4\cdot 4}{x^2}=32.\end{array}$$ Multiplying the first equation by 4, the second equation by $\frac{1}{4}$, and the fourth equation by $\frac{1}{16}$, we obtain an equivalent system $$\{\begin{array}{l}{x}^2+\frac{1}{y^2}=2,\\ y^2+\frac{1}{z^2}=2,\\ z^2+\frac{1}{t^2}=2,\\ t^2+\frac{1}{x^2}=2.\end{array}$$ Adding the equations of this system gives $$x^2+\frac{1}{y^2}+y^2+\frac{1}{z^2}+z^2+\frac{1}{t^2}+t^2+\frac{1}{x^2}=8.$$ But for every real number $u$, $u+\frac{1}{u}\ge 2$, where equality holds only if $u=1$. By adding up inequalities $x^2+\frac{1}{x^2}\ge 2$, $y^2+\frac{1}{y^2}\ge 2$, $z^2+\frac{1}{z^2}\ge 2$ and $t^2+\frac{1}{t^2}\ge 2$, we get $$x^2+\frac{1}{x^2}+y^2+\frac{1}{y^2}+z^2+\frac{1}{z^2}+t^2+\frac{1}{t^2}\ge 8.$$ This inequality must hold as equality by the above; hence all four previous inequalities must hold as equalities, too, i.e., $x^2=y^2=z^2=t^2=1$. As the numbers are positive, the only possibility is $x=y=z=t=1$. Hence $abcd=\frac{x}{2}\cdot 2y\cdot z\cdot 4t=4xyzt=4$.