Đáp án: GTNN M= 49/16
Giải thích các bước giải:
$\begin{array}{l}
M = \frac{1}{{16{x^2}}} + \frac{1}{{4{y^2}}} + \frac{1}{{{z^2}}}\\
= \left( {\frac{1}{{16{x^2}}} + \frac{1}{{4{y^2}}} + \frac{1}{{{z^2}}}} \right).\left( {{x^2} + {y^2} + {z^2}} \right)\left( {do:{x^2} + {y^2} + {z^2} = 1} \right)\\
\ge {\left( {\frac{1}{4} + \frac{1}{2} + 1} \right)^2} = \frac{{49}}{{16}}\left( {Bunhia} \right)\\
\Rightarrow {M_{\min }} = \frac{{49}}{{16}} \Leftrightarrow \left\{ \begin{array}{l}
{x^2} = \frac{1}{7}\\
{y^2} = \frac{2}{7}\\
{z^2} = \frac{4}{7}
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
x = \frac{1}{{\sqrt 7 }}\\
y = \sqrt {\frac{2}{7}} \\
z = \sqrt {\frac{4}{7}}
\end{array} \right.
\end{array}$
