Question

URGENT! a) Sa se arate ca x2 + y2 ≥ 2xy, oricare ar fi numerele x, y ∈ R.

b) Sa se arate ca x2 + y2 + z2 ≥ xy + xz + yz, oricare ar fi x, y, z ∈ R.

Va multumesc anticipat!!

Answer 1

$a) {x}^{2} + {y}^{2} \geqslant 2xy$

${x}^{2} + {y}^{2} - 2xy \geqslant 0$

${x}^{2} - 2xy + {y}^{2} \geqslant 0$

${(x - y)}^{2} \geqslant 0 \: \: \: \forall \: x,y\:\in\:\mathbb{R}$

$b) {x}^{2} + {y}^{2} + {z}^{2} \geqslant xy + xz + yz$

${x}^{2} + {y}^{2} + {z}^{2} - xy - xz - yz \geqslant 0 \: | \times 2$

$2 {x}^{2} + 2 {y}^{2} + 2 {z}^{2} - 2xy - 2xz - 2yz \geqslant 0$

${x}^{2} + {x}^{2} + {y}^{2} + {y}^{2} + {z}^{2} + {z}^{2} - 2xy - 2xz - 2yz \geqslant 0$

${x}^{2} - 2xy + {y}^{2} + {x}^{2} - 2xz + {z}^{2} + {y}^{2} - 2yz + {z}^{2} \geqslant 0$

${(x -y )}^{2} + {(x - z)}^{2} + {(y - z)}^{2} \geqslant 0 \: \: \:\forall \: x,y,z\:\in\:\mathbb{R}$