Comparing $\sqrt{5} + \sqrt{6} + \sqrt{11}$ and $8$ without calculating the values 
I want to compare $\sqrt{5} + \sqrt{6} + \sqrt{11}$ and $8$ without calculating the actual value of square roots.

I tried to apply square on both side but it still carries the root terms.
Any trick I was missing here?

Note that $8>\sqrt{11}$. Therefore\begin{align}\sqrt5+\sqrt6+\sqrt{11}>8&\iff\sqrt5+\sqrt6>8-\sqrt{11}\text{ ($3$ square roots)}\\&\iff11+2\sqrt{30}>75-16\sqrt{11}\text{ (only $2$ square roots)}\\&\iff\sqrt{30}>32-8\sqrt{11}\\&\iff30>1728-512\sqrt{11}\text{ (a single square root)}\\&\iff512\sqrt{11}>1698\\&\iff256\sqrt{11}>849,\end{align}which is true, since $256^2\times11=720\,896$ and $849^2=720\,801$.
