Using AM-GM inequality prove that $(1+\sqrt{2})(1+\sqrt{3})(1+\sqrt{5}) \lt 8\sqrt{30}$.

Question

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?

Answer 1

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

Answer 2

$$(1+\sqrt 2)(1+\sqrt 3)(1+\sqrt 5)=1+\sqrt 2 + \sqrt 3 + \sqrt 5 + \sqrt 6 + \sqrt {10} + \sqrt {15} + \sqrt {30}$$ 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 $$\frac{1+\sqrt 2 + \sqrt 3 + \sqrt 5 + \sqrt 6 + \sqrt {10} + \sqrt {15} + \sqrt {30}}{8}\leq \sqrt{\frac{72}{8}}=3\leq \sqrt{30}$$