Question

Fie x,y,x>0 si x+y+z=3.Aratati ca:
2(x³+y³+z³)≥x²+y²+z²+3
Se rezolva cu ~CEBISEV~! :)

Answer 1

$\displaystyle Tripletele~(x,y,z)~si~(x^2,y^2,z^2)~sunt~la~fel~orientate,~deci \\ \\ conform~inegalitatii~lui~Cebisev,~avem: \\ \\ \sum x^3= \sum (x \cdot x^2) \geq \frac{1}{3} \sum x \cdot \sum x^2= \sum x^2. \\ \\ Deci~ \sum x^3 \geq \sum x^2 \geq \frac{(x+y+z)^2}{3}=3. \\ \\ Am~demonstrat~ca: \\ \\ x^3+y^3+z^3 \geq x^2+y^2+z^2 \\ \\ x^3+y^3+z^3 \geq 3 \\ \\ Prin~insumarea~acestor~doua~relatii~rezulta~concluzia.$

$Am~folosit~bine-cunoscuta~inegalitate~ \\ \\ x^2+y^2+z^2 \geq \frac{(x+y+z)^2}{3}.~(pentru~demonstratie~poti~vedea \\ \\ ultimul~meu~raspuns)$