o inegalitate

[Andi Brojbeanu]

Demonstrati inegalitatea:
\( \frac{x}{13x+y+z}+\frac{y}{13y+x+z}+\frac{z}{13z+x+y}\le \frac{1}{5} \).

[Mateescu Constantin]

Presupunem, fara a restrange generalitatea ca \( x\ge y\ge z \) . Atunci : \( \left\[\ \begin{array}{cc} \frac{x}{13x+y+z} & \ge & \frac{y}{13y+x+z} & \ge & \frac{z}{13z+x+y} \\\\\\\\\\\
13x+y+z & \ge & 13y+x+z & \ge & 13z+x+y\ \end{array}\right\] \) ,

si aplicam inegalitatea lui Cebasev : \( \sum\ \[\frac{x}{13x+y+z}\cdot(13x+y+z)\]\ \ge\ \frac 13\cdot\sum\ \frac{x}{13x+y+z}\cdot\sum\ (13x+y+z) \) .

\( \Longleftrightarrow\ x+y+z\ \ge\ \frac 13\ \cdot\ \(\sum\ \frac{x}{13x+y+z}\)\ \cdot\ 15(x+y+z)\ \Longleftrightarrow\ \sum\ \frac{x}{13x+y+z}\ \le\ \frac 15 \) .

[Claudiu Mindrila]

O solutie la nivelul clasei a VIII-a se poate da notand \( a=13x+y+z, \ b=x+13y+z, \ c=x+y+13z \) , iar apoi exprimand \( x, \ y, \ z \) in functie de \( a, \ b, \ c \).