Sa se arete ca relatia este mai mica ca trei:

[tex]a) \frac{ \sqrt{2} }{3} + \frac{ \sqrt{6} }{5} + \frac{ \sqrt{12} }{7} + \frac{ \sqrt{20} }{9} + \frac{ \sqrt{30} }{11} + \frac{ \sqrt{42} }{13} < 3 \: \: \: (A) \\ \\ \frac{ \sqrt{2} }{3} + \frac{ \sqrt{6} }{5} + \frac{2 \sqrt{3} }{7} + \frac{2 \sqrt{5} }{9} + \frac{ \sqrt{30} }{11} + \frac{ \sqrt{42} }{13} < 3 \\ \\ 2,949526 < 3 [/tex]

[tex]\it Folosim\ inegalitatea\ mediilor:\\ \\ m_g\leq m_a \Rightarrow \sqrt{ab}\leq\dfrac{a+b}{2} \Rightarrow \dfrac{\sqrt{ab}}{a+b}\leq\dfrac{1}{2}\\ \\ \\ \dfrac{\sqrt{2}}{3} = \dfrac{\sqrt{1\cdot2}}{1+2} \leq\dfrac{1}{2}[/tex]

[tex]\it \dfrac{\sqrt{6}}{5} = \dfrac{\sqrt{2\cdot3}}{2+3} \leq\dfrac{1}{2} \\ \\ \\ \it \dfrac{\sqrt{12}}{7} = \dfrac{\sqrt{3\cdot4}}{3+4} \leq\dfrac{1}{2}\\.\\.\\.\\ \it \dfrac{\sqrt{42}}{13} = \dfrac{\sqrt{6\cdot7}}{6+7} \leq\dfrac{1}{2}[/tex]

Notăm membrul stâng cu s și însumând, se obține :

[tex]\it s\leq \underbrace{\it\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}+\ ...\ +\dfrac{1}{2}}_{6\ termeni} \Rightarrow s\leq6\cdot\dfrac{1}{2} \Rightarrow s\leq3[/tex]