Solved with nukelauncher. Let $x=2^a$, $y=2^b$, $z=2^c$ be positive reals. Then the given rewrites as \[(x^2+1)(y^2+2)(z^2+8)=32xyz.\]However observe that \begin{align*} x^2+1&\ge2x\\ y^2+2&\ge2\sqrt2y\\ z^2+8&\ge4\sqrt2z. \end{align*}Multiplying gives \[(x^2+1)(y^2+2)(z^2+8)\ge32xyz,\]with equality iff $(x,y,z)=(1,\sqrt2,2\sqrt2)$. Then the only solution is $(a,b,c)=(0,\tfrac12,\tfrac32)$.

first if we consider rationals then we see that 2^2a + 1 is odd and the rhs is a power of 2 so we see that a has to be zero then we have that (2^2b+2)(2^2c+8)=2^b+c+4 but now we can factor a 2 common from all the equations and we get that 2^2b-1 +1 is odd so we see that 2b is 1 so b is half and so on we get the answer for integers

It took me a while(around $15$ minutes, thinking that it is probably a NT problem) to realize the AM-GM during the test... Anyways, by AM-GM inequality \begin{align*} 2^{2a}+1&\geq 2^{a+1}\\ 2^{2b}+2&\geq 2^{b+\frac{3}{2}}\\ 2^{2c}+8&\geq 2^{c+\frac{5}{2}}\\ \end{align*}Multiplying them we have $L.H.S.\geq R.H.S.$, so all the inequalities must be equalities, so $$a=0,b=\frac{1}{2},c=\frac{3}{2}$$as desired.

TheUltimate123 wrote: Solved with nukelauncher. Let $x=2^a$, $y=2^b$, $z=2^c$ be positive reals. Then the given rewrites as \[(x^2+1)(y^2+2)(z^2+8)=32xyz.\]However observe that \begin{align*} x^2+1&\ge2x\\ y^2+2&\ge2\sqrt2y\\ z^2+8&\ge4\sqrt2z. \end{align*}Multiplying gives \[(x^2+1)(y^2+2)(z^2+8)\ge32xyz,\]with equality iff $(x,y,z)=(1,\sqrt2,2\sqrt2)$. Then the only solution is $(a,b,c)=(0,\tfrac12,\tfrac32)$. Nice solution

This one is a bit tricky, by AM-GM we have that $$2^{2a}+1 \ge 2^{a+1}$$$$2^{2b}+2 \ge 2^{b+\frac{3}{2}}$$$$2^{2c}+8 \ge 2^{c+\frac{5}{2}}$$So now by multiplying all the ineqs we get $(2^{2a}+1)(2^{2b}+2)(2^{2c}+8) \ge 2^{a+b+c+5}$ so since we have the equality we get that $(a,b,c)=\left(0,\frac{1}{2},\frac{3}{2} \right)$ thus we are done

We have: $$(2^{2a}+1)(2^{2b}+2)(2^{2c}+8)\ge2^{a+1}\cdot2^{b+\frac32}\cdot2^{c+\frac52}=2^{a+b+c+5}$$with equality iff $2^{2a}-1=2^{2b}-2=2^{2c}-8=0$, so $(a,b,c)=\left(0,\frac12,\frac32\right)$.

For convenience, let x,y,z equal 2^a,2^b,2^c, respectively, rephrasing it into (x^2+1)(y^2+2)(z^2+8)=32xyz. But by AM-GM on each of the respective terms on LHS and multiplying them, this implies equality must hold; in particular, x=1,y=sqrt2,z=2sqrt2, so (a,b,c)=(0,1/2,3/2) oops reverted to no latex cuz this one is too short

Observe that we have $$2^{2a} + 1 \ge 2\sqrt{2^{2a}} = 2^{a+1}$$$$2^{2b} + 2 \ge 2\sqrt{2^{2b}\cdot{2}} = 2^{b+\frac{3}{2}}$$$$2^{2c} + 8 \ge 2\sqrt{2^{2c}\cdot 8} = 2^{c+ \frac{5}{2}}.$$Multiplying all of these, we get $$(2^{2a}+1)(2^{2b}+2)(2^{2c}+8) \ge 2^{a+b+c+5}.$$Therefore, equality holds in all three cases, so we have $$2^{2a} = 1 \implies a = 0$$$$2^{2b} = 2 \implies b = \frac{1}{2}$$$$2^{2c} = 8 \implies c = \frac{3}{2}.$$Since $(a, b, c) = \left(0, \frac{1}{2}, \frac{3}{2} \right)$ clearly works, it's indeed the final answer. $\square$