Square roots vs. integer

$\color{#D61F06}{\sqrt{2}}>\color{#3D99F6}{1.4}\ (2>1.4^2=1.96)\\ \color{#D61F06}{\sqrt{3}}>\color{#3D99F6}{1.7}\ (3>1.7^2=2.89)\\ \color{#D61F06}{\sqrt{5}}>\color{#3D99F6}{2.2}\ (5>2.2^2=4.84)\\ \color{#D61F06}{\sqrt{6}}>\color{#3D99F6}{2.4}\ (6>2.4^2=5.76)\\ \color{#D61F06}{\sqrt{7}}>\color{#3D99F6}{2.6}\ (7>2.6^2=6.76)\\ \color{#D61F06}{\sqrt{8}}>\color{#3D99F6}{2.9}\ (8>2.8^2=7.84)\\ \implies \color{#D61F06}{\sqrt{2}+\sqrt{3}+\sqrt{5}+\sqrt{6}+\sqrt{7}+\sqrt{8}}>\color{#3D99F6}{1.4+1.7+2.2+2.4+2.6+2.8}=13.1>4+9$ So the answer is $\boxed{>}$

when you square the above 2 sides you will get : $(\sqrt{2}+\sqrt{3}+\sqrt{5}+\sqrt{6}+\sqrt{7}+\sqrt{8})^{2}$ = $(\sqrt{2}+\sqrt{3}+\sqrt{5})^{2}+(\sqrt{6}+\sqrt{7}+\sqrt{8})^{2}+(2×(\sqrt{2}+\sqrt{3}+\sqrt{5})×(\sqrt{6}+\sqrt{7}+\sqrt{8}))=177.04$ which is greater than $(4+9)^{2}=169$ so $\boxed{\geq}$

I will say until the first digit of those irrational numbers. $\sqrt{2} = 1.4, \sqrt{3} = 1.7, \sqrt{5} = 2.2, \sqrt{6} = 2.4, \sqrt{7} = 2.6, \sqrt{8} = 2.8.$ I know those numbers are not equal, but it is natural to calculate easier to use approximate numbers. Then add the numbers. We get $13.1 \boxed{} 13.$ Then the answer is $\boxed{>}.$