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Dacă x, y, z sunt numere reale pozitive, sa se arate ca:
b)
[tex]x + y + z \geqslant \sqrt{xy} + \sqrt{yz} + \sqrt{xz} [/tex]


Răspuns :

Explicație pas cu pas:

Pornim de la :

(Vx-Vy)²>=0

(Vx-Vz)²>=0

(Vy-Vz)²>=0

Adica:

x+y>=2Vxy

x+z>=2Vxz

y+z>=2Vyz

Le adunam:

x+y+x+z+y+z>=2Vxy+2Vxz+2Vyz =>

2x+2y+2z>=2(Vxy+Vxz+Vyz)=>

2(x+y+z)>=2(Vxy+Vxz+Vyz)=>

x+y+z>=Vxy+Vxz+Vyz (q.e.d)

Met 2

x+y+z>=Vxy+Vxz+Vyz /*2 =>

2x+2y+2z>=2Vxy+2Vxz+2Vyz =>

2x+2y+2z-2Vxy-2Vxz-2Vyz>=0 =>

x-2Vxy+y+x-2Vxz+z+y-2Vyz+z>=0 =>

(Vx-Vy)²+(Vx-Vz)²+(Vy-Vz)²>=0 (A) q.e.d