Showing $x+y>z$, where $x=\sqrt{10}+\sqrt{26}$, $y=\sqrt{17}+\sqrt{37}$, and $z=\sqrt{323}$. Is my idea corect?

Question

Let the numbers be $x=\sqrt{10}+\sqrt{26}$, $y=\sqrt{17}+\sqrt{37}$, and $z=\sqrt{323}$. Show that $x+y>z$.

MY IDEA

So I thought if i write every number in a interval of 2 perfect square numbers.

Using this table we can write that $\sqrt{9}<\sqrt{10}<\sqrt{16}$, which basiclly means that $3<\sqrt{10}<4$

We can do this to all numbers and add them up. We can compare them easly after.

I just wanted to ask if my ideas and my way of thinking is corect. If not, can you help me solve this problem? Thank you so much!