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.

$$\begin{eqnarray*}
   N=\sqrt{3}+\sqrt{7}+\sqrt{21} & \Rightarrow & N-\sqrt{21}=\sqrt{3}+\sqrt{7}   \\
    & \Rightarrow & {{N}^{2}}+21-2N\sqrt{21}=10+2\sqrt{21}   \\
    & \Rightarrow& \sqrt{21}=\frac{{{N}^{2}}+11}{2+2N}  \\
\end{eqnarray*}$$
So $\sqrt{21}$ is rational, which is a contradiction
