$p,q,r$ primes, $\sqrt{p}+\sqrt{q}+\sqrt{r}$ is irrational. I want to prove that for $p,q,r$ different primes, $\sqrt{p}+\sqrt{q}+\sqrt{r}$ is irrational. 
Is the following proof correct?
If $\sqrt{p}+\sqrt{q}+\sqrt{r}$ is rational, then  $(\sqrt{p}+\sqrt{q}+\sqrt{r})^2$ is rational, thus $p+q+r+2\sqrt{pq}+2\sqrt{pr}+2\sqrt{qr}$ is rational, therefore $\sqrt{pq}+\sqrt{pr}+\sqrt{qr}$ is rational.
If $\sqrt{pq}+\sqrt{pr}+\sqrt{qr}$ is rational, then $(\sqrt{pq}+\sqrt{pr}+\sqrt{qr})^2$ is rational, therefore $pq+qr+pr+\sqrt{p^2qr}+\sqrt{pq^2r}+\sqrt{pqr^2}$ is rational, therefore $p\sqrt{qr}+q\sqrt{pr}+r\sqrt{pq}$ is rational.
Now suppose $p<q<r$. If $p\sqrt{qr}+q\sqrt{pr}+r\sqrt{pq}$  and $\sqrt{pq}+\sqrt{pr}+\sqrt{qr}$ are rational, then $$p\sqrt{qr}+q\sqrt{pr}+r\sqrt{pq}-p(\sqrt{pq}+\sqrt{pr}+\sqrt{qr})$$
is rational, therefore $(q-p)\sqrt{pr}+(r-p)\sqrt{pq}$ is rational. 
If $(q-p)\sqrt{pr}+(r-p)\sqrt{pq}$ is rational, then $((q-p)\sqrt{pr}+(r-p)\sqrt{pq})^2$ is rational, thus $(q-p)^2pr+2(q-p)(r-p)\sqrt{p^2qr}+(r-p)^2pq$ is rational, thus $\sqrt{qr}$ is rational.
But $q,r$ are distinct primes, thus $qr$ can't be a square. Thus $\sqrt{qr}$ is irrational. Contradiction. 
Also, is there an easier proof?
 A: There is an simplier proof:
Suppose that $\sqrt{p}+\sqrt{q}+\sqrt{r}=w$, where $w$ is rational. Then$\sqrt{p}+\sqrt{q}=-\sqrt{r}+w$, so:
$$(\sqrt{p}+\sqrt{q})^2=(-\sqrt{r}+w)^2$$
Therefore $\sqrt{pq}+w\sqrt{r}$ is rational. But it isn't (the same argument). Suppose that $\sqrt{pq}-w\sqrt{r}=s$ where $s$ is rational, then:
$$\sqrt{pq}=s+w\sqrt{r}$$
$$(\sqrt{pq})^2=(s+w\sqrt{r})^2$$
So $sw\sqrt{r}$ is rational (we also should check case when $s=0$ or $w=0$, but it is simple).
A: Your proof is actually very close to valid.  But in the end, you require that $p, q, r$ be distinct primes, and that is a stronger condition than in the problem as posed.
However, the proof you give is very easy to make complete.  For example, if $p = r \neq q$ then we have to show $2\sqrt{p} + \sqrt{q}$ is irrational, and for that the squaring proof works easily.  And if $p=q=r$ we have to show that $3\sqrt{p}$ is irrational, again trivial.
A: A tricky proof may be the following: assume that $\sqrt{p}+\sqrt{q}+\sqrt{r}$ is a rational number $\frac{a}{b}$ with $\gcd(a,b)=1$. Then for every prime $s>b$ we have that the following relation about Legendre symbols:
$$ \left(\frac{p}{s}\right)=\left(\frac{q}{s}\right)=1 $$
implies $\left(\frac{r}{s}\right)=1$. Let $\eta_r$ be the least quadratic non-residue $\!\!\pmod{r}$. 
The Chinese remainder theorem and the Dirichlet's theorem give that there are an infinite number of primes $s$ for which $s\equiv 1\pmod{4pq}$ and $s\equiv \eta_r\pmod{r}$, but quadratic reciprocity then gives:
$$ \left(\frac{p}{s}\right)=\left(\frac{q}{s}\right)=1,\quad \left(\frac{r}{s}\right)=\left(\frac{\eta_r}{r}\right)=-1, $$
contradiction.
A: because $(\sqrt{p}+\sqrt{q})(\sqrt{p}-\sqrt{q}) \in \mathbb{Q}$ we have 
$$
\sqrt{p} + \sqrt{q} \in  \mathbb{Q}(\sqrt{r}) \Rightarrow \sqrt{p} - \sqrt{q} \in  \mathbb{Q}(\sqrt{r}) \Rightarrow \sqrt{p} \in  \mathbb{Q}(\sqrt{r})
$$
so for some $a,b \in \mathbb{Q}$ with $ab\ne 0$
$$
\sqrt{p} = a+ b \sqrt{r} 
$$
squaring both sides shows this cannot be the case, so the assumption $\sqrt{p} + \sqrt{q} \in  \mathbb{Q}(\sqrt{r})$ must be at fault. in particular we cannot have:
$$
\sqrt{p}+\sqrt{q}+\sqrt{r} \in \mathbb{Q}
$$
A: Suppose $\sqrt{p}+\sqrt{q}+\sqrt{r} = a \to (a-\sqrt{p})^2 = (\sqrt{q}+\sqrt{r})^2\to a^2-2a\sqrt{p} +p =q+r+2\sqrt{qr}\to a^2-q+p-r=2a\sqrt{p}+2\sqrt{qr}\to (a^2-q+p-r)^2=4a^2p+4qr+8a\sqrt{pqr}\to \sqrt{pqr} $ is a rational number, and this is not possible.
