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Let $x = \sqrt{6 + \sqrt{5}}$ and $y = \sqrt{5 + \sqrt{6}}.$ Since both $x$ and $y$ are positive, it will suffice to compare $x^{2}$ and $y^{2}.$

Now $x^{2} - y^{2} = 6 + \sqrt{5} - (5 + \sqrt{6}) = 1 - (\sqrt{6} - \sqrt{5}) \gt 1 - (\sqrt{9} - \sqrt{4}) = 1 - 1 = 0.$

Thus $x^{2} \gt y^{2},$ implying that $x \gt y,$ i.e., the greater value is $\boxed{\sqrt{6 + \sqrt{5}}}.$

$\large \color{#302B94}{\sqrt{6+\sqrt{5}} = \sqrt{5+\sqrt{5}+1 }} \\ \large \color{#302B94}{\sqrt{5+\sqrt{5}+1 }} ~~~\boxed{?}~~~ \color{#69047E}{\sqrt{5+\sqrt{6} } }$

So all we need to find out now is which is greater, $\sqrt{5}+1$ or $\sqrt{6}$ . Square both sides and the answer will become clear:

$\large \sqrt{5}+1 ~~~\boxed{?}~~~ \sqrt{6} \\ \large {(\sqrt{5}+1)}^{2} ~~~\boxed{?}~~~ {(\sqrt{6})}^{2} \\ \large 5+2\sqrt{5}+1 ~~~\boxed{?}~~~ 6 \\ \large 6+2\sqrt{5} ~~~\boxed{>}~~~ 6$

Hence $\large \sqrt{6+\sqrt{5}} > \sqrt{5+\sqrt{6}}$ , so our answer is $\large \color{#20A900}{\boxed{\sqrt{6+\sqrt{5}}}}$ .

Since 3 > $\sqrt5$ > 2, and 3 > $\sqrt6$ > 2:

6 + 2+ > 8 while 5 + 2+ < 8.

Obviously, $\sqrt{6 + \sqrt5}$ > $\sqrt{5 + \sqrt6}$ .

Answer: $\boxed{\sqrt{6 + \sqrt5}}$