Old AMO problem

By AM - GM inequality , we have 2 cases:

$\boxed{1.-}$ Let's suppose $d \ge 1$ , then $4(abc)^{\frac{1}{2}} \leq 4(abcd)^{\frac{1}{2}} \leq a^2 + b^2 + c^2 + d^2 \Rightarrow$ $256(abc)^{2} \leq 256(abcd)^{2} \leq (a^2 + b^2 + c^2 + d^2)^{4}.$ And ( $256(abcd)^{2} = (a^2 + b^2 + c^2 + d^2)^{4} \iff a = b = c = d$ ). In this case,if ( $a = b = c = d$ ) and if $d \ge 2, \space \text{ then } (abc)^2 < (abcd)^2 = a^8$ (this case lead us to one case which is impossible)

and if $d =1 = a = b = c, (abc)^2 = 1 < 1^2 + 1^2 + 1^2 + 1^2 = 4$ (this case is also impossible), so the only possibility is $d = 0$ .

$\boxed{2.-}$ If $d = 0$ by AM - GM inequality again, $3(abc)^{\frac{2}{3}} \leq a^2 + b^2 + c^2 \Rightarrow (abc)^2 \leq 27(abc)^2 \leq (a^2 + b^2 + c^2)^3$ and the last inequality is only an equality if and only if $a = b = c$ . In this case, $(abc)^2 = a^6 \leq 27a^6$ and hence, the equation $(abc)^2 = a^2 + b^2 + c^2$ has the only solution $(a,b,c) = (0,0,0)$ and all this implies that the only solution is $(a,b,c,d) = (0,0,0,0)$