For all positive integer $x\geq1$, show $\left\lfloor\sqrt{x}+\sqrt {x+1}+\sqrt {x+2}\right\rfloor=\left\lfloor\sqrt{9x+8}\right\rfloor$

Question

This is the first problem on the CMO Qualifying Repêchage this year, and I don't know how to solve it. Any help would be greatly appreciated.

Show that for all integers $x\geq1$,

$$\left\lfloor\sqrt{x}+\sqrt{x+1}+\sqrt{x+2}\right\rfloor = \left\lfloor \sqrt{9x+8}\right\rfloor$$

My first attempt was to prove it by induction. I tried to show that if

$$p\leq\left\lfloor\sqrt{x}+\sqrt{x+1}+\sqrt{x+2}\right\rfloor < p+1 $$ and

$$p\leq\left\lfloor \sqrt{9x+8}\right\rfloor<p+1$$

then

$$q\leq\left\lfloor\sqrt{x+1}+\sqrt{x+2}+\sqrt{x+3}\right\rfloor < q+1 $$ and

$$q\leq\left\lfloor \sqrt{9(x+1)+8}\right\rfloor<q+1$$ for integers p and q, But I wasn't able to make it work.

Then I tried to prove that the LHS was greater than the RHS for all $$x\geq1$$ and that.difference between LHS and RHS was less than the fractional part of the LHS, but this didn't work either. At this point, I'm not sure how else to attack the problem. Any help would be greatly appreciated.

This is the link to the original problem set: https://cms.math.ca/Competitions/REP/2020/2020CMOQR_exam_en.pdf