Question

Daca x,y > 0, atunci: $\it 2(\frac{x^{2} }{y^{2} } + \frac{y^{2} }{x^{2} } )$ ≥ $\it \frac{x}{y} + \frac{y}{x} + 2.$

MULTUMESC!!

Answer 1

Răspuns:

Explicație pas cu pas:

Answer 2

$\it \forall\ x,\ y>0 \Rightarrow \dfrac{x}{y}+\dfrac{y}{x} \geq2\ \ \ \ (1)\\ \\ \\ 2\Big(\dfrac{x^2}{y^2}+\dfrac{y^2}{x^2}\Big)\geq \dfrac{x}{y}+\dfrac{y}{x}+2 \Leftrightarrow 2\Big[\Big(\dfrac{x}{y}+\dfrac{y}{x}\Big)^2-2\Big] \geq \dfrac{x}{y}+\dfrac{y}{x}+2\ \ \ \ \ (2)\\ \\ \\ Vom\ nota\ \dfrac{x}{y}+\dfrac{y}{x} = t \stackrel{(1)}{\Longrightarrow}\ t\geq2\ \ \ (3)\\ \\ \\ Inegalitatea\ (2)\ devine:$

$\it 2(t^2-2)\geq t+2 \Leftrightarrow 2t^2-4\geq t+2 \Leftrightarrow 2t^2-t\geq 2+4 \Leftrightarrow \\ \\ \Leftrightarrow t(2t-1)\geq6\ \ (Adev\breve{a}rat,\ \ \forall\ t\geq2)$