If any two or more of the variables are zero, equality clearly holds. If exactly one of the variables is zero, say WLOG $t=0$, we need to prove $$\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\geq 3\sqrt[3]{xyz}$$which is directly true by AM-GM. So, we now assume that all the variables are positive. Since the inequality is homogeneous, let $t=1$. For no specific reason, replace $(x,y,z)$ by $(a^2,b^2,c^2)$ Let $a+b+c=3u, ab+bc+ca=3v^2, abc=w^3$ We need to prove $$u+v^2\geq\sqrt[3]{9v^4-6uw^3+w^6}$$Cubing, it is equivalent to proving $$u^3+(v^6-w^6)+3u^2v^2+3uv^4+6uw^3-9v^4\geq 0$$Since $v^6\geq w^6$ is true, we need to prove $$u^3+3u^2v^2+3uv^4+6uw^3-9v^4\geq 0$$The expression is linear and increasing in $w^3$, by Tejs theorem it suffices to check for the case $a=c$, which gives $$6 a^5 + 27 a^4 b + 11 a^4 + 36 a^3 b^2 + 36 a^3 b + 8 a^3 + 12 a^2 b^3 + 25 a^2 b^2 + 12 a^2 b + 6 a b^3 + 6 a b^2 + b^3\geq 0$$clearly true. No equality holds in this case since $a,b,c$ are positive reals. Hence, equality holds when atleast two of the variables are zero, or $(x,x,x,0)$ and it's permutations.