Điều kiện :$x,y \ge \frac 58$
$\ggg\sqrt{24(x^2+y^2+4)}=\Biggl(x\sqrt{8y-5}+y\sqrt{8x-5} \Biggl)^2 \overset{BCS}\le \big(x^2+y^2 \big) \Bigg(8(x+y)-10 \Bigg) $
$\le \big( x^2+y^2 \big)\Bigg(8\sqrt{2(x^2+y^2)}-10 \Bigg)\Leftrightarrow x^2+y^2 \ge 2$
$\ggg 0=11x^2-6xy+3y^2-2x+4y \ge 9x^2-6xy+3y^2-2x+4y+4=(-3x+y+2)^2$
$\Leftrightarrow \begin{cases}-3x+y+2= 0\\ x^2+y^2=2 \end{cases}\Leftrightarrow \color{red}{\boxed{\begin{cases}x= 1\\ y= 1\end{cases}}}$