Question

demonstrati ca √2+√6+√12+√20+√30<17,5

Answer 1

$Aplicam~inegalitatea~mediilor:~\sqrt{ab} \leq \frac{a+b}{2} . \\ \\ \sqrt{2}+\sqrt{6}+\sqrt{12}+\sqrt{20}+\sqrt{30}= \\ \\ =\sqrt{1 \cdot 2}+ \sqrt{2 \cdot 3 }+ \sqrt{3 \cdot 4}+ \sqrt{4 \cdot 5}+ \sqrt{5 \cdot 6}\ \textless \ \\ \\ \ \textless \ \frac{1+2}{2}+ \frac{2+3}{2} + \frac{3+4}{2}+ \frac{4+5}{2} + \frac{5+6}{2} = \frac{35}{2}=17,5.$