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: Suppose $(1+\sqrt3)(1+\sqrt7)=p/q$ for some $p,q\in\Bbb Z^+$. Then we have that $$q(1+\sqrt3)=\frac p{1+\sqrt7}=\frac{p(1-\sqrt7)}{-6}\implies p\sqrt7-6q\sqrt3=p+6q\ne0\tag1$$ This implies that $$p\sqrt7+6q\sqrt3=\frac{(p\sqrt7+6q\sqrt3)(p\sqrt7-6q\sqrt3)}{p\sqrt7-6q\sqrt3}=\frac{7p^2-108q^2}{p+6q}\tag2$$ Adding $(1)$ and $(2)$ together gives $$2p\sqrt7=p+6q+\frac{7p^2-108q^2}{p+6q}\implies\sqrt7\in\Bbb Q$$ which is a contradiction. $\square$
A: If $(1+\sqrt{3})(1+\sqrt{7})$ is rational, then 
$$\displaystyle \frac{12}{(1+\sqrt{3})(1+\sqrt{7})}=\frac{12(1-\sqrt{3})(1-\sqrt{7})}{(-2)(-6)}=1-\sqrt{3}-\sqrt{7}+\sqrt{21}$$ is also rational. 
So, $\displaystyle \frac{1}{2}[(1+\sqrt{3})(1+\sqrt{7})+1-\sqrt{3}-\sqrt{7}+\sqrt{21}]-1=\sqrt{21}$ is rational.
This leads to a contradiction.
A: $$\begin{eqnarray*}
   N=\sqrt{3}+\sqrt{7}+\sqrt{21} & \Rightarrow & N-\sqrt{21}=\sqrt{3}+\sqrt{7}   \\
    & \Rightarrow & {{N}^{2}}+21-2N\sqrt{21}=10+2\sqrt{21}   \\
    & \Rightarrow& \sqrt{21}=\frac{{{N}^{2}}+11}{2+2N}  \\
\end{eqnarray*}$$
So $\sqrt{21}$ is rational, which is a contradiction
A: 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 ...
A: Hint $\,\sqrt{21}+\sqrt{7}+\sqrt{3}=q\,\Rightarrow\ \sqrt7(\sqrt 3+1) = q-\sqrt3  \,\Rightarrow\, \sqrt 7\in \Bbb Q(\sqrt3)$, contradiction as below.
Lemma $\rm\ \ [K(\sqrt{a},\sqrt{b}) : K] = 4\ $ if  $\rm\ \sqrt{a},\ \sqrt{b},\ \sqrt{a\,b}\ $  all are not in $\rm\,K\,$ and $\rm\, 2 \ne 0\,$ in the field $\rm\,K.$
Proof $\  $  Let  $\rm\ L = K(\sqrt{b}).\,$ Then $\rm\,  [L:K] = 2\,$  via  $\rm\,\sqrt{b}  \not\in K,\,$  so it suffices to prove $\rm\, [L(\sqrt{a}):L] = 2.\,$ It fails only if  $\rm\,\sqrt{a} \in L = K(\sqrt{b})\, $ and then $\rm\, \sqrt{a}\ =\  r + s\, \sqrt{b}\ $  for $\rm\ r,s\in K.\,$ But that's impossible,
since squaring  $\Rightarrow \rm(1)\!:\ \ a\ =\ r^2 + b\ s^2 + 2\,r\,s\  \sqrt{b},\, $ which contradicts hypotheses as follows:
$\rm\qquad\qquad rs \ne 0\ \ \Rightarrow\ \  \sqrt{b}\ \in\  K\ \ $ by solving $(1)$ for $\rm\sqrt{b}\,,\,$ using  $\rm\,2 \ne 0$
$\rm\qquad\qquad\  s = 0\ \ \Rightarrow\ \  \ \sqrt{a}\ \in\  K\ \ $  via  $\rm\ \sqrt{a}\ =\ r+s\,\sqrt b = r \in K$
$\rm\qquad\qquad\  r = 0\ \ \Rightarrow\ \  \sqrt{a\,b}\in K\ \ $  via  $\rm\ \sqrt{a}\ =\ s\, \sqrt{b},\ \ $times $\rm\,\sqrt{b}\quad$
Remark $ $  The Lemma generalizes to any number of sqrts). See the citations there for generalizations to $n$'th roots.
A: In this answer I proved the  theorem below about positive sums of square roots in ordered fields.
Theorem $\ \sqrt{c_1}+\cdots+\!\sqrt{c_{n}} \in K\ \Rightarrow \sqrt{c_i}\in K\,$ for all $\, i,\:$ if $\,0 < c_i\in K$  an ordered field.
It is instructive to specialize that proof here. Readers unfamiliar with fields should begin here.
Assuming  $\, q =  \sqrt 3\! +\!\sqrt 7\!+\!\sqrt{21} \in\Bbb Q\,$ we infer a $\color{#0a0}{\text{contradiction: one of }\,\sqrt3,\sqrt 7,\sqrt{21}\,\ {\rm is} \in \Bbb Q}$.
By the Lemma below: $\ \sqrt{7} + \sqrt{21} = q\!-\!\sqrt 3 \in \Bbb Q(\sqrt 3)\,\Rightarrow\, \sqrt{7}, \sqrt{21} \in \Bbb Q(\sqrt 3),\,$ thus
$$\begin{align} \sqrt{7}\, &= \,a\, +\, b\sqrt 3,\ \ \ \ a,\,b\ \in\Bbb Q\\[.3em] \sqrt{21}\, &=\, a' + b'\sqrt 3,\ \ \ a',b'\in\Bbb Q\end{align}\qquad\quad\ \ $$
If $\, b\, < 0\,$ then $\,a = \sqrt 7 - b\sqrt 3 = \sqrt 7 + \sqrt{3b^2}\in\Bbb Q\Rightarrow\color{#0a0}{\sqrt 7\in \Bbb Q}\,$ by the Lemma.
If $\ b'\! < 0\,$ then the same argument allows us to deduce  $\color{#0a0}{\sqrt{21}\in \Bbb Q}$
Else $\,b,b'\ge 0\,\Rightarrow\,\color{#c00}{1\!+\!b\!+\!b' > 0}\,$ so, by below, we infer $\, \color{#0a0}{\sqrt 3\,\in \Bbb Q},\,$ a $\rm\color{#0a0}{\rm contradiction}$ in all cases.
$\quad\ \ q = \sqrt 3\! +\!\sqrt 7\!+\!\sqrt{21} = a\!+\!a'+(1\!+\!b\!+\!b')\sqrt 3\,\ $ so $\ \sqrt 3 = \dfrac{q\!-\!a\!-\!a'}{\color{#c00}{1\!+\!b\!+\!b'}}\in\Bbb Q$
Lemma $ $ If $\,0< r,s\in K$ then $\, k = \sqrt r\! +\!\sqrt s\in K\Rightarrow\sqrt r,\sqrt s\in K,\,$ for any subfield $\,K\subseteq \Bbb R$
Proof $\ $ Note $\ k' = \sqrt{r}\!-\!\sqrt{s}\: = \dfrac{\ \ r\, -\ s}{\sqrt{r}\!+\!\sqrt{s}}\in K\ $  by $\,0 < \sqrt r\! +\! \sqrt s\in K,\,$ by $\, \sqrt r,\sqrt s > 0$
Therefore $\ (k+k')/2 = \sqrt r\in K\ $ and $\ (k-k')/2 = \sqrt s\in K$.
Remark $ $ Above is the inductive step of the general proof specialized to the case $\,P(2)\Rightarrow P(3),\,$ where $\,P(n)\,$ denotes the proposition with a sum of $\,n\,$ square roots. The general induction step works precisely the same way.
