$p,q,r$ 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?

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.
