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?

Use the Taylor expansions:
$$\sqrt{1+x}=\sqrt{x} + \frac1{2\sqrt x} - \frac1{8x\sqrt x}+\frac1{16x^2\sqrt x}-\frac5{128x^3\sqrt x}+O(x^{-4})\\
\sqrt{2+x}=\sqrt{x} + \frac1{\sqrt x} - \frac1{2x\sqrt x}+\frac1{2x^2\sqrt x}-\frac5{8x^3\sqrt x}+O(x^{-4})$$
Now we will plug in and calculate (manually possible, though tedious):
$$2.23606...=\sqrt{1+4}>2+\frac1{4}-\frac1{64}+\frac1{512}-\frac5{16384}\approx 2.23602;\\
2.4494...=\sqrt{2+4}>2+\frac12-\frac1{16}+\frac1{64}-\frac5{1024}\approx 2.4482;\\
3.3166...=\sqrt{2+9}>3+\frac13-\frac1{54}+\frac1{486}-\frac5{17496}\approx 3.3165.$$
Hence:
$$\sqrt5+\sqrt6+\sqrt{11}>8.00062>8$$
