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?
 A: 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 $849^2=720\,801$.
A: We'll prove that:
$$\sqrt5+\sqrt6>8-\sqrt{11}$$ or
$$11+2\sqrt{30}>64-16\sqrt{11}+11$$ or
$$\sqrt{15}+4\sqrt{22}>16\sqrt2$$ or
$$15+352+8\sqrt{15\cdot22}>512$$ or
$$8\sqrt{15\cdot22}>145$$ or
$$8\sqrt{66}>29\sqrt5$$ or
$$64\cdot66>841\cdot5$$ or
$$65^2-1>4205$$ or
$$4225-1>4205,$$ which is true.
A: 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.
A: 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:
$$\sqrt5+\sqrt6+\sqrt{11}>8.00062>8$$
