we have: $ \sqrt {2} < 2$ and $ \sqrt {6} < \sqrt {9}$ so $ \sqrt {2 + \sqrt {2 + \sqrt {2 + ... + \sqrt {2}}}} < \sqrt {2 + ...\sqrt {2 + \sqrt {2 + 2}}} = 2$ and ${ \sqrt {6 + \sqrt {6 + \sqrt {6 + ... + \sqrt {6}}}} < \sqrt {6 + ...\sqrt {6 + \sqrt {6 + 3}}}} = 3$ after it's obvious

The following inequality is also true. For every positive integer $ n$ prove the inequality $ \sqrt{1+\sqrt{3+\sqrt{5+\sqrt{\cdots+\sqrt{2n-1}}}}}<2$. (Ukraine)

This one needs a variant of induction I suppose... "Backward Induction".We show the general statement $\sqrt{2m-1+\sqrt{...+\sqrt{2n-1}}}<m+1$ is true for $m+1$ and then show for $m$.So let the expression be $f(m)$ and we have $f(m+1)<m+2$ from the inductive step.Now $2m-1+f(m+1)<3m+1 \implies [f(m)]^2<3m+1\le (m+1)^2$.thus proved.The last step is valid since $m\ge 1\implies m^2\ge m\implies m^2+2m+1\ge 3m+1$ .

Prove the inequality ${\sqrt{3}<\sqrt{1+\sqrt{{2+\sqrt{3+\ldots }}}}}< \sqrt{1+\sqrt{5}}.$ Prove the inequality $ \sqrt{2012+\sqrt{2011+\sqrt{\cdots+\sqrt{2+\sqrt{1}}}}}<46$. (Hungary 2012--2013 ) Pumac 2013:Evaluate$\sqrt{2013+276\sqrt{2027+278\sqrt{2041+280\sqrt{2055+\cdots}}}}$. Generalization of Ukraine 1997 here

Let $f_k(n)=\underbrace{\sqrt{k+\sqrt{k+\sqrt{k+\ldots+\sqrt k}}}}_{n\text{ times}}$. $\textbf{1. }f_2(n)<2$ Easy to see that $\sqrt2<2$. If $f_2(n)<2$ for some $n$, then $f_2(n+1)=\sqrt{f_2(n)+2}<\sqrt{2+2}=2$, and induction finishes. $\blacksquare$ $\textbf{2. }f_6(n)<3$ Similarly, $\sqrt6<3$. Then $f_6(n)<3\Rightarrow f_6(n+1)=\sqrt{f_6(n)+6}<\sqrt{3+6}=3$. $\blacksquare$ Adding these produces the desired inequality. $\square$