Using the fact that $\sqrt{n}$ is an irrational number whenever $n$ is not a perfect square, show $\sqrt{3} + \sqrt{7} + \sqrt{21}$ is irrational. 
Question:
Using the fact that $\sqrt{n}$ is an irrational number whenever $n$ is not a perfect square, show $\sqrt{3} + \sqrt{7} + \sqrt{21}$ is irrational.

Following from the question, I tried:

Let $N = \sqrt{3} + \sqrt{7} + \sqrt{21}$. Then,
$$
\begin{align}
N+1 &= 1+\sqrt{3} + \sqrt{7} + \sqrt{21}\\
    &= 1+\sqrt{3} + \sqrt{7} + \sqrt{3}\sqrt{7}\\
    &= (1+\sqrt{3})(1+\sqrt{7}).
\end{align}
$$
Using the above stated fact, $\sqrt{3}$ and $\sqrt{7}$ are irrational. Also, sum of a rational and irrational number is always irrational, so $1+\sqrt{3}$ and $1+\sqrt{7}$ are irrational. Similarly, if we prove that $N+1$ is irrational, $N$ will also be proved to be irrational.

But, how do I prove that product of $1+\sqrt{3}$ and $1+\sqrt{7}$ are irrational.

A somewhat systematic (but laborious) approach: Assume
$$N=\sqrt 3+\sqrt 7+\sqrt{21} $$
is rational. Then also 
$$N^2=3+7+21+2(\sqrt{21}+3\sqrt 7+7\sqrt 3)= 31+2\sqrt{21}+3\sqrt 7+7\sqrt 3$$
is rational, as well as
$$(N^2-31)^2 =4\cdot 21+9\cdot 7+49\cdot 3+2(42\sqrt 3+42\sqrt 7+21\sqrt{21}).$$
Thus also
$$(N^2-31)^2- (4\cdot 21+9\cdot 7+49\cdot 3)-84N=-42\sqrt{21}$$
is rational.
I guess you can see how this could be similarly applied to all specific sums of square roots ...
