Question

Sa se demonstreze:
(x+y+z)^2<=3(x^2+y^2+z^2)

Answer 1

(x+y+z)²≤3(x²+y²+z²)

x²+y²+z²≤3x²+3y²+3z²-2xy-2yz-2xz

x²+y²+z²≤(x²-2xy+y²)+(y²-2yz+z²)+(x²-2xz+z²)+x²+y²+z²

0≤(x-y)²+(y-z)²+(x-y)²

sau  (x-y)²+(y-z)²+(x-y)²≥0;    adevarat, suma a trei patrate ≥0

=>(x+y+z)²≤3(x²+y²+z²)