Chủ đề này có 1 trả lời

[DOTOANNANG]

$\lceil$ COME ! ! ! BACK $\rfloor$ $x,\,y,\,z> 0$

$$x\,\frac{\sqrt{2\,x(\,x+ y\,)}}{z(\,z+ x\,)}+ y\,\frac{\sqrt{2\,y(\,y+ z\,)}}{x(\,x+ y\,)}+ z\,\frac{\sqrt{2\,z(\,z+ x\,)}}{y(\,y+ z\,)}\geqq 3$$

@HaiDangel²⁽⁰¹⁾⁹

 

[DOTOANNANG]

Sử dụng $\lceil$ HOLDER!inequality $\rfloor$ ta chứng minh

$$x\,\frac{\sqrt{2\,x(\,x+ y\,)}}{z(\,z+ x\,)}+ y\,\frac{\sqrt{2\,y(\,y+ z\,)}}{x(\,x+ y\,)}+ z\,\frac{\sqrt{2\,z(\,z+ x\,)}}{y(\,y+ z\,)}\geqq 3$$

$$\because\,\sqrt{\dfrac{x(\,x+ y\,)}{\frac{z^{\,2}(\,z+ x\,)^{\,2}}{x^{\,2}}}}+ \sqrt{\dfrac{y(\,y+ z\,)}{\frac{x^{\,2}(\,x+ y\,)^{\,2}}{y^{\,2}}}}+ \sqrt{\dfrac{z(\,z+ x\,)}{\frac{y^{\,2}(\,y+ z\,)^{\,2}}{z^{\,2}}}}\geqq \frac{3}{\sqrt{2}}$$

$$\because\,\dfrac{\left ( x(\,x+ y\,)+ y(\,y+ z\,)+ z(\,z+ x\,) \right )^{\,3}}{(\,x+ y\,)^{\,2}z^{\,2}(\,z+ x\,)^{\,2}+ (\,y+ z\,)^{\,2}x^{\,2}(\,x+ y\,)^{\,2}+ (\,z+ x\,)^{\,2}y^{\,2}(\,y+ z\,)^{\,2}}\geqq \frac{9}{2}$$

@HaiDangel

$\lceil$ https://math.stackex.../3249550/679470 $\rfloor$ [on hold]

Sử dụng $\lceil$ A.M.*G.M.!inequality $\rfloor$ ta chứng minh

$$\frac{4\,x^{\,2}(\,x+ y\,)}{2\,z(\,z+ x\,)\sqrt{2\,x(\,x+ y\,)}}+ \frac{4\,y^{\,2}(\,y+ z\,)}{2\,x(\,x+ y\,)\sqrt{2\,y(\,y+ z\,)}}+ \frac{4\,z^{\,2}(\,z+ x\,)}{2\,y(\,y+ z\,)\sqrt{2\,z(\,z+ x\,)}}\geqq 3$$

$$\because\,\frac{4\,x^{\,2}(\,x+ y\,)}{z(\,z+ x\,)(\,3\,x+ y\,)}+ \frac{4\,y^{\,2}(\,y+ z\,)}{x(\,x+ y\,)(\,3\,y+ z\,)}+ \frac{4\,z^{\,2}(\,z+ x\,)}{y(\,y+ z\,)(\,3\,z+ x\,)}\geqq 3$$

 Michael Rozenberg