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?

The following is a somewhat more laborious, but also more generic, way to solve the problem.

*

*The minimal (rational) polynomial of $\,x_0=\sqrt{5}+\sqrt{6}+\sqrt{11}\,$ is
$$
P(x)=x^8 - 88 x^6 + 1696 x^4 - 10560 x^2 + 14400
$$
which can be determined similar to Minimal Polynomial of $\sqrt{2}+\sqrt{3}+\sqrt{5}$.


*With the substitution $\,x \to x+4\,$:
$$P(x+4) = x^8 + 32 x^7 + 360 x^6 + 1472 x^5 - 1504 x^4 - 28160 x^3 - 70976 x^2 - 59904 x - 15296
$$
By Descartes' rule of signs $A(x)=P(x+4)$ has exactly one real positive root, since there is one change of signs and $A(0) \lt 0\,$. It follows that $P(x)$ has exactly one real root larger than $\,4\,$ and that root must be $\,x_0\,$ since $\,x_0=\sqrt{5}+\sqrt{6}+\sqrt{11} \gt \sqrt{4}+\sqrt{4} = 4\,$.


*With the substitution $\,x \to x+8\,$:
$$P(x+8) = x^8 + 64 x^7 + 1704 x^6 + 24448 x^5 + 203936 x^4 + 988160 x^3 + 2574016 x^2 + 2780160 x - 6080
$$
Again by Descartes' rule of signs $B(x)=P(x+8)$ has exactly one real positive root, since there is one change of signs and $B(0) \lt 0\,$. It follows that $P(x)$ has exactly one real root larger than $\,8\,$, and that root must be the unique root larger than $\,4\,$ as established at the previous step, so in the end $\,x_0 \gt 8\,$.
As a side note, the middle step and related calculations could have been avoided by noting that the roots of $\,P(x)\,$ are the rational conjugates $\,\pm\sqrt{5}\pm\sqrt{6}\pm\sqrt{11}\,$, so $\,x_0\,$ is the largest real root.
