Giải thích các bước giải:
Ta có:
$M=\dfrac{1}{16x^2}+\dfrac{1}{4y^2}+\dfrac{1}{z^2}$
$\to M=\dfrac{(\dfrac14)^2}{x^2}+\dfrac{(\dfrac12)^2}{y^2}+\dfrac{1^2}{z^2}$
$\to M\ge \dfrac{(\dfrac14+\dfrac12+1)^2}{x^2+y^2+z^2}$
$\to M\ge \dfrac{(\dfrac14+\dfrac12+1)^2}{1}$ vì $x^2+y^2+z^2=1$
$\to M\ge\dfrac{49}{16}$
Dấu = xảy ra khi:
$\dfrac{\dfrac14}{x^2}=\dfrac{\dfrac12}{y^2}=\dfrac1{z^2}=\dfrac{\dfrac14+\dfrac12+1}{x^2+y^2+z^2}=\dfrac74$
$\to\dfrac{\dfrac14}{x^2}=\dfrac{\dfrac12}{y^2}=\dfrac1{z^2}=\dfrac74$
$\to x^2=\dfrac17, y^2=\dfrac27, z^2=\dfrac47$
$\to x=\sqrt{\dfrac17}, y=\sqrt{\dfrac27}, z=\sqrt{\dfrac47}$
