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Prove the inequality $$ \sqrt {2 + \sqrt [3]{3 + ... + \sqrt [{1993}]{1993}}} < 2$$ All-Russian MO 1993 Regional (Round 4) 10.6

Generalization: $\sqrt{2+\sqrt[3]{3+...+\sqrt[n]{n}}}<2$

$\sqrt[n]{n+2} \leq 2$ for $n \geq 2$ Easy to prove by induction $\sqrt{2+\sqrt[3]{3+...+\sqrt[n]{n}}}<\sqrt{2+\sqrt[3]{3+...+\sqrt[n]{n+2}}} \leq \sqrt{2+\sqrt[3]{3+...+\sqrt[n-1]{n-1+2}}} \leq... \leq \sqrt{2+2}=2$