Cho các số thực không âm \(x,y,z\) thỏa mãn:

Ta chứng minh bổ đề: \(\frac{1}{{{m^2}}} + \frac{1}{{{n^2}}} \ge \frac{8}{{{{(m + n)}^2}}}\)

Áp dụng BĐT Cô si: \(\frac{1}{{{m^2}}} + \frac{1}{{{n^2}}} \ge 2\sqrt {\frac{1}{{{m^2}{n^2}}}}  = \frac{2}{{mn}}.\,\,\)

\(Do\,\,mn \le \frac{{{{(m + n)}^2}}}{4}\)   Nên suy ra \(\frac{1}{{{m^2}}} + \frac{1}{{{n^2}}} \ge \frac{8}{{{{(m + n)}^2}}}\)

  Ta có:

 \[\begin{array}{l}4{{\rm{x}}^2} + 4 \ge 2\sqrt {4{{\rm{x}}^2}.4}  = 8{\rm{x}}\\4{y^2} + 4 \ge 8y\\4{z^2} + 4 \ge 8{\rm{z}}\end{array}\]

Cộng vế với vế ta được: \(24 \ge 8x + 8z + 2y \Leftrightarrow 3 \ge {\rm{x}} + {\rm{z}} + \frac{y}{4}\)

Ta lại có:

\(\frac{1}{{{{(z + 1)}^2}}} + \frac{{16}}{{{{(y + 4)}^2}}} + \frac{8}{{{{(x + 3)}^2}}} = \frac{1}{{{{(z + 1)}^2}}} + \frac{1}{{{{\left( {\frac{y}{4} + 1} \right)}^2}}} + \frac{8}{{{{(x + 3)}^2}}} \ge \frac{8}{{{{\left( {z + \frac{y}{4} + 2} \right)}^2}}} + \frac{8}{{{{(x + 3)}^2}}} \ge \)

\( \ge 8.\frac{8}{{{{\left( {z + \frac{y}{4} + 2 + x + 3} \right)}^2}}} \ge 8.\frac{8}{{{{(3 + 2 + 3)}^2}}} = 1\)

\(\frac{1}{{{{(z + 1)}^2}}} + \frac{{16}}{{{{(y + 4)}^2}}} + \frac{8}{{{{(x + 3)}^2}}} + 2023 \ge 1 + 2023 = 2024\)

Dấu “=” xảy ra khi \(x = 1;y = 4;z = 1\)