Using AM-GM inequality prove that $(1+\sqrt{2})(1+\sqrt{3})(1+\sqrt{5}) \lt 8\sqrt{30}$. It is trivial to prove that $(1+\sqrt{2})(1+\sqrt{3})(1+\sqrt{5}) \le 
8\sqrt{30}$  using numeric methods. For example by multiplying
$(1+\sqrt{2}) \le  3 $
$(1+\sqrt{3}) \le  3 $
$(1+\sqrt{5}) \le  4 $
We get:
$(1+\sqrt{2})(1+\sqrt{3})(1+\sqrt{5}) < 36$
while $8\sqrt{30} \gt 40 $ because $\sqrt{30} \gt 5 $ . However for this 
particular problem it is asked a solution using the AM-GM inequality and I am not able to find one. Can anyone help me?

We use the inequality $$\frac{a+b}{2}\le \sqrt{\frac{a^2+b^2}{2}}$$ so we get
$$\frac{1+\sqrt{2}}{2}\times\frac{1+\sqrt{3}}{2}\times\frac{1+\sqrt{5}}{2}\le \sqrt{30}$$
