<s>[INST] 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?
 [/INST] $$(1+\sqrt 2)(1+\sqrt 3)(1+\sqrt 5)=1+\sqrt 2 + \sqrt 3 + \sqrt 5 + \sqrt 6 + \sqrt {10} + \sqrt {15} + \sqrt {30}$$
and thus you can use the AM-GM inequality on this
Actually I think I messed up, the AM-GM inequality goes the wrong way and the exponents don't work out. Use the Root-Mean Square - Arithmetic Mean Inequality: https://artofproblemsolving.com/wiki/index.php/Root-Mean_Square-Arithmetic_Mean-Geometric_Mean-Harmonic_mean_Inequality
to get that 
$$\frac{1+\sqrt 2 + \sqrt 3 + \sqrt 5 + \sqrt 6 + \sqrt {10} + \sqrt {15} + \sqrt {30}}{8}\leq \sqrt{\frac{72}{8}}=3\leq \sqrt{30}$$
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