Show that $\sqrt{abc}$ is irrational if $a, b, c$, and $\sqrt{a} + \sqrt{b} + \sqrt{c}$ are irrational.

Question

Assume $a$,$b$,$c$ are the irrational numbers, and $\sqrt a + \sqrt b + \sqrt c$ is irrational number. Show that $\sqrt{abc}$ is irrational number. Please help me this problem, thank you for watching!

Answer 1

The statement is not true. However, the counterexamples in the comments don't work. Note the requirement that $a, b,$ and $c$ (not just $\sqrt{a}$, $\sqrt{b}$, and $\sqrt{c}$) should be irrational.

For a correct counterexample, take $$ a = b = c = \sqrt[3]{4}. $$ These are irrational (this can be proven similarly to the irrationality of the square root of $2$). Then, $$ \sqrt{a} + \sqrt{b} + \sqrt{c} = \sqrt[3]{2} + \sqrt[3]{2} + \sqrt[3]{2} = 3\sqrt[3]{2}, $$ which is also irrational -- because $\sqrt[3]{2}$ is irrational (which can again be proven similarly to irrationality of the square root of $2$).

However, $$ \sqrt{abc} = \sqrt{4} = 2, $$ which is rational.