a) Aplicăm inegalitatea mediilor:
[tex]\it m_g\leq m_a[/tex]
[tex]\it \sqrt{a_1a_2} \leq \dfrac{1}{2}(a_1+a_2) \Rightarrow a_1a_2\leq \dfrac{1}{4} (a_1+a_2)^2 \Rightarrow \\\;\\ \\\;\\ \Rightarrow \dfrac{a_1a_2}{a_1+a_2} \leq\dfrac{ \dfrac{1}{4} (a_1+a_2)^2}{a_1+a_2} \Rightarrow \dfrac{a_1a_2}{a_1+a_2} \leq \dfrac{a_1+a_2}{4} [/tex]
Acum, membrul stâng al inegalității din enunț devine:
[tex]\it \dfrac{a_1a_2}{a_1+a_2} +\dfrac{a_2a_3}{a_2+a_3}+ ... +\dfrac{a_na_1}{a_n+a_1} \leq \dfrac{a_1+a_2+a_2+a_3+ ... +a_n+a_1}{4}
[/tex]
[tex]\it \Rightarrow \dfrac{a_1a_2}{a_1+a_2} +\dfrac{a_2a_3}{a_2+a_3}+ ... +\dfrac{a_na_1}{a_n+a_1} \leq \dfrac{2(a_1+a_2+a_3+ ... +a_n)}{4}\Rightarrow[/tex]
[tex]\it \Rightarrow \dfrac{a_1a_2}{a_1+a_2} +\dfrac{a_2a_3}{a_2+a_3}+ ... +\dfrac{a_na_1}{a_n+a_1} \leq \dfrac{2\cdot1}{4} =\dfrac{1}{2}[/tex]
