Exercice 8 Montrer que : ∀(x,y)∈R2 : 2x2+xy+y2≥∣x−y∣\forall(x,y)\in\R^2~:~2\sqrt{x^2+xy+y^2} \ge |x-y|∀(x,y)∈R2 : 2x2+xy+y2≥∣x−y∣ Correction Soit x,y∈Rx,y\in\Rx,y∈R on a : 2x2+xy+y2≥∣x−y∣ ⟺ (2x2+xy+y2)2≥∣x−y∣2 ⟺ 4(x2+xy+y2)≥x2−2xy+y2 ⟺ 3x2+3y2+6xy≥0 ⟺ x2+2xy+y2≥0 ⟺ (x+y)2≥0 \begin{align*} &2\sqrt{x^2+xy+y^2} \ge |x-y| \\ & \iff \left(2\sqrt{x^2+xy+y^2}\right)^2 \ge |x-y|^2 \\ & \iff 4(x^2+xy+y^2) \ge x^2-2xy+y^2 \\ & \iff 3x^2+3y^2+6xy \ge 0 \\ & \iff x^2+2xy+y^2 \ge 0 \\ & \iff (x+y)^2 \ge 0 \\ \end{align*}2x2+xy+y2≥∣x−y∣⟺(2x2+xy+y2)2≥∣x−y∣2⟺4(x2+xy+y2)≥x2−2xy+y2⟺3x2+3y2+6xy≥0⟺x2+2xy+y2≥0⟺(x+y)2≥0 On sait que (x+y)2≥0(x+y)^2 \ge 0(x+y)2≥0 (vraie) Donc : ∀(x,y)∈R2 : 2x2+xy+y2≥∣x−y∣\forall(x,y)\in\R^2~:~2\sqrt{x^2+xy+y^2} \ge |x-y|∀(x,y)∈R2 : 2x2+xy+y2≥∣x−y∣