Question

Fie a1,a2,a3, ..... ,an numere pozitive care au suma egala cu 1

Aratati ca:
a1*a2 a2*a3 ....... an*a1
------- + -------- + + -------<sau=1/2
a1+a2 a2+a3 an+a1

Answer 1

$\text{Aplicam inegalitatea dintre media geometrica si media aritmetica:}\\\boxed{\sqrt{x\cdot y}\leq \dfrac{x+y}{2}}, \text{ sau }  \boxed{x\cdot y\leq \dfrac{(x+y)^2}{4}}\\\text{Avem ca :}\\\dfrac{a_k\cdot a_{k+1} }{a_k+a_{k+1}}\leq \dfrac{(a_k+a_{k+1})^2}{4(a_4+a_{k+1})} = \dfrac{a_k+a_{k+1}}{4}\\\text{Aplicand pentru fiecare fractie in parte obtinem :}$

$\dfrac{a_1\cdot a_2}{a_1+a_2}+\dfrac{a_2\cdot a_3}{a_2+a_3}+\ldots +\dfrac{a_n\cdot a_1}{a_n+a_1}\leq \dfrac{a_1+a_2 }{4}+\dfrac{a_2+a_3}{4}+\ldots +\dfrac{a_n+a_1}{4}=\\~~~~~~~~~~~~~~\leq 1\\\dfrac{2\overbrace{(a_1+a_2+\ldots +a_n)}}{4}\leq \dfrac{2}{4}=\dfrac{1}{2}~ Q.E.D.$