Question

Demonstrati ca x/y + y/z + z/x $\geq$ 3

Answer 1

Răspuns:

Conf inegalitatii mediilor

(x/y+y/z+z/x)/3≥∛(x/y*y/z*z/x)

(x/y+y/z+z/x)/3≥∛1

(x/y+y/z+z/x)/3≥1

x/y+y/z+z/x≥3

Explicație pas cu pas:

Answer 2

$\dfrac{x}{y}+\dfrac{y}{z}+\dfrac{z}{x} \geq 3$

Inegalitatea mediilor:

$\left.\begin{cases}\dfrac{x}{y}+\dfrac{y}{z}\geq 2\sqrt{\dfrac{x}{y}\cdot \dfrac{y}{z}}\\ \dfrac{z}{x}+1 \geq 2\sqrt{\dfrac{z}{x}\cdot 1}\end{cases}\right|\Rightarrow \dfrac{x}{y}+\dfrac{y}{z}+\dfrac{z}{x}+1 \geq 2\sqrt{\dfrac{x}{z}}+2\sqrt{\dfrac{z}{x}}$

$\Rightarrow \dfrac{x}{y}+\dfrac{y}{z}+\dfrac{z}{x}+1 \geq 2\left(\dfrac{\sqrt{x}}{\sqrt{z}}+\dfrac{\sqrt{z}}{\sqrt{x}}\right)$

$\dfrac{\sqrt{x}}{\sqrt{z}}+\dfrac{\sqrt{z}}{\sqrt{x}}\geq 2\sqrt{\dfrac{\sqrt{x}}{\sqrt{z}}\cdot \dfrac{\sqrt{z}}{\sqrt{x}}} = 2\sqrt{1} = 2$

$\Rightarrow \dfrac{x}{y}+\dfrac{y}{z}+\dfrac{z}{x}+1 \geq 2\cdot 2$

$\Rightarrow \boxed{\dfrac{x}{y}+\dfrac{y}{z}+\dfrac{z}{x} \geq 3}$