Cho ba số thực dương x, y, z thỏa mãn \(4xy + 2yz + 3xz = 24\). Tìm giá trị lớn nhất của biểu thức \[P = \frac{{2x}}{{\sqrt {{x^2} + 4} }} + \frac{y}{{\sqrt {{y^2} + 9} }} + \frac{z}{{\sqrt {{z^2} + 16} }}\].

Ta có: \(4xy + 2yz + 3xz = 24 \Leftrightarrow \frac{{xy}}{6} = \frac{{yz}}{{12}} = \frac{{xz}}{8} = 1\)\( \Leftrightarrow \frac{x}{2}.\frac{y}{3} + \frac{y}{3}.\frac{z}{4} + \frac{x}{2}.\frac{z}{4} = 1\)         

Đặt \(\frac{x}{2} = a > 0;\frac{y}{3} = b > 0;\frac{z}{4} = c > 0 \Rightarrow ab + bc + ac = 1\)

\(\begin{array}{l}P = \frac{{4a}}{{\sqrt {4{a^2} + 4} }} + \frac{{3b}}{{\sqrt {9{b^2} + 9} }} + \frac{{4c}}{{\sqrt {16{c^2} + 16} }}\\ = \frac{{2a}}{{\sqrt {{a^2} + 1} }} + \frac{b}{{\sqrt {{b^2} + 1} }} + \frac{c}{{\sqrt {{c^2} + 1} }}\end{array}\)

\( = \frac{{2a}}{{\sqrt {{a^2} + ab + bc + ca} }} + \frac{b}{{\sqrt {{b^2} + ab + bc + ac} }} + \frac{c}{{\sqrt {{c^2} + ab + bc + ac} }}\)

\( = \frac{{2a}}{{\sqrt {\left( {a + b} \right)\left( {a + c} \right)} }} + \frac{b}{{\sqrt {\left( {a + b} \right)\left( {b + c} \right)} }} + \frac{c}{{\sqrt {\left( {a + c} \right)\left( {b + c} \right)} }}\)

\( = \sqrt {\frac{{2a}}{{a + b}}.\frac{{2a}}{{a + c}}}  + \sqrt {\frac{{2b}}{{a + b}}.\frac{b}{{2\left( {b + c} \right)}}}  + \sqrt {\frac{c}{{2\left( {b + c} \right)}}.\frac{{2c}}{{a + c}}} \)

Ta có :

\(\begin{array}{l}\frac{{2a}}{{a + b}} + \frac{{2a}}{{a + c}} \ge 2\sqrt {\frac{{2a}}{{a + b}}.\frac{{2a}}{{a + c}}} \\\frac{{2b}}{{a + b}} + \frac{b}{{2\left( {b + c} \right)}} \ge 2\sqrt {\frac{{2b}}{{a + b}}.\frac{b}{{2\left( {b + c} \right)}}} \\\frac{c}{{2\left( {b + c} \right)}} + \frac{{2c}}{{a + c}} \ge 2\sqrt {\frac{c}{{2\left( {b + c} \right)}}.\frac{{2c}}{{a + c}}} \end{array}\)

\(p \le \frac{1}{2}\left( {\frac{{2a}}{{a + b}} + \frac{{2a}}{{a + c}} + \frac{{2b}}{{a + b}} + \frac{b}{{2\left( {b + c} \right)}} + \frac{c}{{2\left( {b + c} \right)}} + \frac{{2c}}{{a + c}}} \right)\)

  \(\begin{array}{l} \Leftrightarrow P \le \frac{1}{2}\left( {\frac{{2\left( {a + b} \right)}}{{a + b}} + \frac{{2\left( {a + c} \right)}}{{a + c}} + \frac{{b + c}}{{2(b + c)}}} \right)\\ \Leftrightarrow P \le \frac{1}{2}\left( {2 + 2 + \frac{1}{2}} \right)\\ \Leftrightarrow P \le \frac{9}{4}\end{array}\)

Dấu  xảy ra khi và chỉ khi \(\left\{ \begin{array}{l}\frac{{2a}}{{a + b}} = \frac{{2a}}{{a + c}}\\\frac{{2b}}{{a + b}} = \frac{b}{{2\left( {b + c} \right)}}\\\frac{c}{{2\left( {b + c} \right)}} = \frac{{2c}}{{a + c}}\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}b = c\\a + b = 8b\\a + c = 8c\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}b = c\\a = 7b\\a = 7c\end{array} \right.\)

\(ab + bc + ac = 1 \Leftrightarrow 7{b^2} + {b^2} + 7{b^2} = 1 \Leftrightarrow {b^2} = \frac{1}{{15}} \Leftrightarrow b = \frac{1}{{\sqrt {15} }}\)

\(\left\{ \begin{array}{l}b = \frac{1}{{\sqrt {15} }} \Rightarrow y = \frac{3}{{\sqrt {15} }}\\c = \frac{1}{{\sqrt {15} }} \Rightarrow z = \frac{4}{{\sqrt {15} }}\\a = \frac{7}{{\sqrt {15} }} \Rightarrow x = \frac{{14}}{{\sqrt {15} }}\end{array} \right.\).